Question 1

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

Accepted Answer

The answer of An object is thrown vertically upward such...

Question 2

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

Accepted Answer

Let's first calculate the maximum height that the ball reaches before starting to fall back down. We can use the formula:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the maximum height we're looking for).

0^2 = 20^2 + 2(-9.8)s
s = 20.4 m

So the maximum height that the ball reaches is 20.4 m.

Now let's look at the stone. We want to find the time it takes for the stone to reach the height of 20.4 m. We can use the formula:

s = ut + (1/2)at^2

where s is the displacement (20.4 m), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time we're looking for.

20.4 = 24t + (1/2)(-9.8)t^2
t = 2.05 s

So it takes the stone 2.05 s to reach the height of 20.4 m.

During this time, the ball has been in the air for a total of 2 + 2.05 = 4.05 s. We can use the formula:

s = ut + (1/2)at^2

to find how far the ball has fallen during this time. We'll take upward to be positive, so a = -9.8 m/s^2.

s = 20(4.05) + (1/2)(-9.8)(4.05)^2
s = 82.4 m

So the ball has fallen 82.4 m during the 4.05 s that the stone was in the air.

Now we can add up the maximum height of the ball and the height that the ball has fallen to find the height of the ball when it passes the stone:

height = 20.4 m + 82.4 m = 102.8 m

Since the question is asking for the height above the release point, we subtract the initial height of the ball:

height above release point = 102.8 m - 20 m = 82.8 m

However, note that this height is above the release point of the ball and not the release point of the stone (which is also at 20 m). Therefore, we need to subtract the height that the stone has traveled during the 2.05 s it took to reach the height of the ball. We can use the same formula as before, but with u = 24 m/s and t = 2.05 s:

s = 24(2.05) + (1/2)(-9.8)(2.05)^2
s = 24.9 m

So the stone has traveled 24.9 m since it was released. Therefore, the height above the release point of the stone is:

height above release point = 82.8 m - 24.9 m = 17.9 m

Rounded to the nearest whole number, the answer is 18 m, which corresponds to choice (C).

Question 3

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

Accepted Answer

Let's denote the initial velocity as "u" and the maximum height reached as "H".
When the object reaches one fourth of its maximum height, it is at a height of H/4. At this point, we can use the equation:

v^2 = u^2 + 2gh

where v is the final velocity (which is 0 when the object reaches maximum height), g is the acceleration due to gravity (-9.8 m/s^2), and h is the height above the launch point. Substituting the given values, we get:

0^2 = u^2 + 2(-9.8)(H/4)

Simplifying, we get:

u^2 = 4.9H

Now, we also know that at the highest point, the velocity is 0. Using the equation for vertical motion under gravity, we get:

v = u + gt

0 = u + (-9.8)t

t = u/9.8

And, we also know that the object takes the same amount of time to reach one fourth of its maximum height and to reach the highest point. So, the time taken to reach H/4 is also u/9.8.

Now, using the equation for vertical displacement, we get:

H/4 = ut + (1/2)gt^2

Substituting the values, we get:

H/4 = u(u/9.8) + (1/2)(-9.8)(u/9.8)^2

Simplifying, we get:

H/4 = u^2/19.6

u^2 = 4.9H

Multiplying both sides by 4, we get:

4u^2 = 19.6H

Substituting the value of H/4 from the previous equation, we get:

4u^2 = 19.6(4u^2)

Solving, we get:

u = 20.5 m/s

Rounding off to one decimal, we get:

u ≈ 21 m/s

Therefore, the answer is D.

Question 4

A ball thrown vertically from ground level is caught 3.0 s later when it is at its highest point by a person on a balcony which is 14 m above the ground. Determine the initial speed of the ball.

Accepted Answer

The ball reaches its highest point in 3.0 s, which means it took 1.5 s to reach the maximum height (since the time to go up equals the time to come down in projectile motion). Using the formula $v = u + at$ for vertical motion (where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration due to gravity (-9.8 m/s$^2$), and $t$ is the time), and knowing that at the highest point the final velocity ($v$) is 0 m/s, we can rearrange the formula to solve for $u$: $u = v - at$. Substituting $v = 0$, $a = -9.8$ m/s$^2$ (taking upward direction as positive), and $t = 1.5$ s, we get $u = 0 - (-9.8 \times 1.5) = 14.7$ m/s. However, this initial speed is only for the vertical component to reach the balcony. To find the total initial speed to reach the balcony at 14 m in 3.0 s, we use the formula for the displacement in the vertical direction: $s = ut + \frac{1}{2}at^2$, where $s$ is the displacement (14 m), $u$ is the initial velocity we're trying to find, $a$ is the acceleration due to gravity (-9.8 m/s$^2$), and $t$ is the time (3.0 s). Rearranging and solving for $u$ gives us the initial speed required to reach 14 m in height over 3.0 s, which correctly matches the scenario described. The calculation provided initially was incorrect in the context of the question's specifics, leading to confusion. The correct approach involves understanding that the total time of flight is given and the height at which the ball is caught. Given these conditions, the initial speed can be found using kinematic equations that relate time, acceleration, and displacement. The correct initial speed, considering the time to reach the highest point and the height of the balcony, aligns with the principles of projectile motion, where the initial speed required to reach a certain height in a given time can be calculated. My explanation mistakenly mixed concepts and did not accurately apply the correct kinematic equation for this scenario. The correct calculation involves using the given height and time to determine the initial velocity needed to reach that height in the specified time, considering the acceleration due to gravity.

Question 5

A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.

Accepted Answer

Using the kinematic equation: 
h = vi*t + 0.5*a*t^2, where 
h = initial height above ground (unknown)
vi = initial velocity = 10 m/s (given)
t = time taken to hit ground = 3.0 s (given)
a = acceleration due to gravity = 9.81 m/s^2
Plugging in the values: 
h = 10*3.0 + 0.5*(-9.81)*(3.0)^2 = 74.145 m
Rounding to the nearest whole number gives the answer of 74 m.

Question 6

The graph below shows the velocity versus time graph for a ball. Which explanation best fits the motion of the ball as shown by the graph?

Accepted Answer

The answer of The graph below shows the velocity versus...

Question 7

The position of a particle moving along the y axis has a position given by  Is there any time interval during which the particle is not moving?

Accepted Answer

The answer of The position of a particle moving along...

Question 8

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s². At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

Accepted Answer

The answer of A rocket, initially at rest, is fired...

Question 9

An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

Accepted Answer

The answer of An object is thrown downward with an...

Question 10

A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s² for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?

Accepted Answer

The rocket accelerates at 20 m/s² for 6 seconds, reaching a velocity of 120 m/s (v = at). The height during this phase is 360 m (s = 0.5 * a * t²). After the motor stops, it continues upward until its velocity is 0, which takes 12 seconds (v = u + at, where u = 120 m/s and a = -9.8 m/s²). The additional height is 720 m (s = ut + 0.5 * at²). Total height = 360 m + 720 m = 1080 m or 1.1 km.

Question 11

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Accepted Answer

The answer of A stone is thrown from the top...

Question 12

When Jim and Rob ride bicycles, Jim can only accelerate at three quarters the acceleration of Rob. Both start from rest at the bottom of a long straight road with constant upward slope. If Rob takes 5.0 minutes to reach the top, how much earlier should Jim start to reach the top at the same time as Rob?

Accepted Answer

Let's assume the slope of the road to be 'a' and the distance between the bottom and top of the road to be 'd'. The time taken by Rob to reach the top of the road can be found using the formula:

$d = \frac{1}{2}at^2$

Using this formula, we get:

$d = \frac{1}{2}a(5*60)^2$

Solving this equation, we get:

$d = 45000a$

Now, let's assume the initial speed of Jim to be 'v'. Using the given information, we can say that his acceleration will be 3/4 times that of Rob. So, his acceleration can be represented as:

$a_j = \frac{3}{4}a_r$

And, using the same formula as before, we can say that:

$d = vt + \frac{1}{2}a_jt^2$

Replacing the value of $a_j$ in this equation, we get:

$d = vt + \frac{3}{8}at^2$

Now, we need to find the time taken by Jim to reach the top. Let's call this time 't_j'. So, we get the following equation for Jim:

$d = vt_j + \frac{3}{8}at_j^2$

We want Jim to reach the top at the same time as Rob. So, we can equate the two equations to get:

$vt_j + \frac{3}{8}at_j^2 = 45000a$

Simplifying this equation, we get:

$t_j^2 + \frac{8v}{3a}t_j - \frac{240000}{a} = 0$

Using the quadratic formula, we get:

$t_j = \frac{-\frac{8v}{3a} \pm \sqrt{\left(\frac{8v}{3a}\right)^2 + 4\left(\frac{240000}{a}\right)}}{2}$

Simplifying this equation, we get:

$t_j = \frac{-4v}{3a} \pm \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}$

Since we want Jim to reach the top at the same time as Rob, we need to choose the positive value of $t_j$. Also, we can ignore the negative sign inside the square root since it will lead to a negative time value. So, we get:

$t_j = \frac{-4v}{3a} + \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}$

Now, we need to find the value of 'v' so that Jim reaches the top at the same time as Rob. We can use the fact that both Jim and Rob start from rest to get this value:

When Jim starts, he needs to cover a distance of 'd' in the time 't_j'. So, using the equation we got earlier:

$d = \frac{1}{2}a_jt_j^2$

Replacing the value of $a_j$, we get:

$d = \frac{3}{8}at_j^2$

Now, we can substitute the value of $t_j$ in this equation to get:

$d = \frac{3}{8}a\left(\frac{-4v}{3a} + \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}\right)^2$

Simplifying this equation, we get:

$v = \frac{3a}{4}\left(1 - \sqrt{1 - \frac{3200}{27ad}}\right)$

Note that we used the given time taken by Rob to calculate the distance 'd' using the formula we used earlier. Substituting the values we get, we get:

$v = 5.49$ m/s

Finally, we need to find the time difference between Rob and Jim:
 
$t_{diff} = t_j - 5.0*60$

Substituting the values we get, we get:

$t_{diff} = 46$ s
 
Therefore, the answer is (C) 46 s.

Question 13

A particle moving along the x axis has a position given by x = 54t − 2.0t³ m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?

Accepted Answer

The given position function is $x = 54t - 2.0t^3$ meters. The speed of the particle is the magnitude of the velocity, which is the first derivative of the position with respect to time, $v = dx/dt = 54 - 6t^2$. Setting this equal to zero to find when the speed is zero gives $54 - 6t^2 = 0$, which has a real solution, indicating the particle does stop at some point (t = 3s as given), but it does not imply the particle remains at rest after t = 3.0 s, nor does it imply the particle no longer accelerates after t = 3.0 s (since acceleration, the second derivative of position, is constant and non-zero, $a = d^2x/dt^2 = -12t$). The particle can be found at positions x < 0 for values of t that make $54t - 2.0t^3 < 0$, which is not limited to t < 0 s. Therefore, none of the given statements are correct.

Question 14

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A little later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:

Accepted Answer

At the instant ball B passes ball A, both balls are at the same position on the hill, meaning they are subject to the same slope and gravitational forces, thus they have the same acceleration. However, since ball B was given a shove and passes ball A, their velocities are different at the passing moment.

Question 15

The position of an object at equal time intervals is shown below:  Which graph below correctly represents position versus time for this object?

Accepted Answer

The answer of The position of an object at equal...

Question 16

When starting from rest at the bottom of a straight road with constant upward slope, Joan bicycles to the top 50.0 s ahead of Sally, whose travel time is 5.00 minutes. What is the ratio of Joan's acceleration to Sally's acceleration?

Accepted Answer

Let's call the distance to the top of the slope "d". 
Sally's average speed is: 
v = d / t = d / 300 s

Joan's average speed is: 
v = d / (t - 50s)

Since we know that Joan started from rest and we want to find the ratio of her acceleration to Sally's acceleration, it's useful to use the kinematic equation: 
d = 1/2 * a * t^2

For Sally: 
d = 1/2 * a_sally * (300 s)^2 
a_sally = 2d / (300 s)^2

For Joan: 
d = 1/2 * a_joan * (t-50s)^2 
2d / (t-50s)^2 = a_joan 
Notice that we can substitute (d / (t-50s)) for Joan's average speed, which gives: 
2 * (d / (t-50s)) / (t-50s) = a_joan

Now we can find the ratio of their accelerations: 
a_joan / a_sally = [2 * (d / (t-50s)) / (t-50s)] / [2d / (300 s)^2] 
a_joan / a_sally = [300^2 / (t-50s)^2] * [(d / (t-50s)) / d] 
a_joan / a_sally = [300^2 / (t-50s)^2] * (1 / (t-50s))

Plugging in the values: 
a_joan / a_sally = [300^2 / (250s)^2] * (1 / 250s) 
a_joan / a_sally = 1.44

Question 17

The velocity at the midway point of a ball able to reach a height y when thrown with velocity v_i at the origin is:

Accepted Answer

The answer of The velocity at the midway point of...

Question 18

To help Kim practice for the Special Olympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use t_Kim for Kim's total time.

Accepted Answer

The answer of To help Kim practice for the Special...

Question 19

A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:

Accepted Answer

The velocity in the horizontal direction remains constant at 10 m/s since there are no forces acting in the horizontal direction after the boy leaves the bench. The acceleration is due to gravity, which is 9.8 m/s^2 downward.

Question 20

Two identical balls are at rest side by side at the bottom of a hill. Some time after ball A is kicked up the hill, ball B is given a kick up the hill. Ball A is headed downhill when it passes ball B headed up the hill. At the instant when ball A passes ball B,

Accepted Answer

At the instant when ball A passes ball B, both balls are at the same position on the hill, and since they are on the same slope, they experience the same acceleration due to gravity. However, their velocities are different because they are moving in opposite directions.

Quiz 2: Motion in One Dimension

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

A ball thrown vertically from ground level is caught 3.0 s later when it is at its highest point by a person on a balcony which is 14 m above the ground. Determine the initial speed of the ball.

A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.

The graph below shows the velocity versus time graph for a ball. Which explanation best fits the motion of the ball as shown by the graph?

The position of a particle moving along the y axis has a position given by Is there any time interval during which the particle is not moving?

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s². At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0) , a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s² for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

A particle moving along the x axis has a position given by x = 54t − 2.0t³ m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?

The position of an object at equal time intervals is shown below: Which graph below correctly represents position versus time for this object?

When starting from rest at the bottom of a straight road with constant upward slope, Joan bicycles to the top 50.0 s ahead of Sally, whose travel time is 5.00 minutes. What is the ratio of Joan's acceleration to Sally's acceleration?

The velocity at the midway point of a ball able to reach a height y when thrown with velocity v_i at the origin is:

To help Kim practice for the Special Olympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use t_Kim for Kim's total time.

A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:

Two identical balls are at rest side by side at the bottom of a hill. Some time after ball A is kicked up the hill, ball B is given a kick up the hill. Ball A is headed downhill when it passes ball B headed up the hill. At the instant when ball A passes ball B,

Quiz 2: Motion in One Dimension

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

A ball thrown vertically from ground level is caught 3.0 s later when it is at its highest point by a person on a balcony which is 14 m above the ground. Determine the initial speed of the ball.

A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.

The graph below shows the velocity versus time graph for a ball. Which explanation best fits the motion of the ball as shown by the graph?

The position of a particle moving along the y axis has a position given by Is there any time interval during which the particle is not moving?

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0) , a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s2 for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

A particle moving along the x axis has a position given by x = 54t − 2.0t3 m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?

The position of an object at equal time intervals is shown below: Which graph below correctly represents position versus time for this object?

When starting from rest at the bottom of a straight road with constant upward slope, Joan bicycles to the top 50.0 s ahead of Sally, whose travel time is 5.00 minutes. What is the ratio of Joan's acceleration to Sally's acceleration?

The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin is:

To help Kim practice for the Special Olympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use tKim for Kim's total time.

A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:

Two identical balls are at rest side by side at the bottom of a hill. Some time after ball A is kicked up the hill, ball B is given a kick up the hill. Ball A is headed downhill when it passes ball B headed up the hill. At the instant when ball A passes ball B,

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s². At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s² for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?

A particle moving along the x axis has a position given by x = 54t − 2.0t³ m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?

The velocity at the midway point of a ball able to reach a height y when thrown with velocity v_i at the origin is:

To help Kim practice for the Special Olympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use t_Kim for Kim's total time.