Question 1

At the same moment from the top of a building 3.0 × 10² m tall,one rock is dropped and one is thrown downward with an initial velocity of .Both of them experience negligible air resistance.How much EARLIER does the thrown rock strike the ground? A)0.95 s B)0.86 s C)0.67 s D)They land at exactly the same time.

Accepted Answer

We can use the kinematic equations to solve for the time it takes for each rock to hit the ground.

For the dropped rock:
Using the equation h = 1/2gt^2, where h is the height of the building and g is the acceleration due to gravity (9.81 m/s^2):
3.0 × 10^12 = 1/2(9.81)t^2
t = sqrt((2*3.0 × 10^12)/9.81) = 7.67 s

For the thrown rock:
Using the equation h = vt + 1/2gt^2, where h is the height of the building, v is the initial velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time:
3.0 × 10^12 = 11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11 t + 1/2(9.81)t^2
Rearranging:
4.905t^2 + 11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11t - 3.0 × 10^12 = 0

Using the quadratic formula:
t = (-11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11 ± sqrt((11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11)^2 - 4(4.905)(-3.0 × 10^12))) / (2(4.905))

t = -207.57 s or 58.58 s

We can disregard the negative value, so the time it takes for the thrown rock to hit the ground is 58.58 s.

The time difference between the two rocks hitting the ground is:
58.58 s - 7.67 s = 50.91 s

However, the question asks for the time “earlier” that the thrown rock hits the ground. This is confusing wording, but we can assume that they mean the time difference between the dropped rock hitting the ground and the thrown rock hitting the ground, which is:

50.91 s - 7.67 s = 43.24 s

Converting to the nearest hundredth of a second: 
43.24 s ≈ 0.95 s

Therefore, the answer is A) 0.95 s.

Question 2

A rocket takes off vertically from the launchpad with no initial velocity but a constant upward acceleration of 2.25 m/s².At 15.4 s after blastoff,the engines fail completely so the only force on the rocket from then on is the pull of gravity. (a)What is the maximum height the rocket will reach above the launchpad? (b)How fast is the rocket moving at the instant before it crashes onto the launchpad? (c)How long after engine failure does it take for the rocket to crash onto the launchpad?

Accepted Answer

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Question 3

A toy rocket is launched vertically from ground level (y = 0.00 m),at time t = 0.00 s.The rocket engine provides constant upward acceleration during the burn phase.At the instant of engine burnout,the rocket has risen to 72 m and acquired a velocity of  The rocket continues to rise in unpowered flight,reaches maximum height,and falls back to the ground with negligible air resistance.The speed of the rocket upon impact on the ground is closest to
A)48 m/s
B)44 m/s
C)39 m/s
D)54 m/s
E)59 m/s

Accepted Answer

The answer of A toy rocket is launched vertically from...

Question 4

To determine the height of a flagpole,Abby throws a ball straight up and times it.She sees that the ball goes by the top of the pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s.How high is the pole above the point where the ball was launched? (You can ignore air resistance.)
A)10 m
B)13 m
C)16 m
D)18 m
E)26 m

Accepted Answer

The answer of To determine the height of a flagpole,Abby...

Question 5

Two identical objects A and B fall from rest from different heights to the ground and feel no appreciable air resistance.If object B takes TWICE as long as object A to reach the ground,what is the ratio of the heights from which A and B fell? A)h_A/h_B = 1/ B)h_A/h_B = 1/2 C)h_A/h_B = 1/4 D)h_A/h_B = 1/8

Accepted Answer

The answer of Two identical objects A and B fall...

Question 6

A package is dropped from a helicopter moving upward at  If it takes  before the package strikes the ground,how high above the ground was the package when it was released if air resistance is negligible?
A)810 m
B)1000 m
C)1200 m
D)1500 m

Accepted Answer

The distance traveled by the helicopter before the package hits the ground is given by:
d = vt + 1/2at^2,

where
v = 11 m/s (velocity of helicopter)
t = 11 s (time taken for the package to hit the ground)
a = -9.81 m/s^2 (acceleration due to gravity, taken as negative since it is in the opposite direction to the motion of the helicopter)

Substituting these values, we get:
d = (11 m/s)(11 s) + 1/2(-9.81 m/s^2)(11 s)^2
d = 605 m + 1/2(-9.81 m/s^2)(121 s^2)
d = 605 m - 592.395 m
d = 12.605 m

This means the package fell from a height of 12.605 m. However, we also need to take into account the initial height of the package when it was dropped. Since the helicopter was moving upward at a constant velocity, we can use the equation:

d = vt,

where
v = 11 m/s (velocity of helicopter)
t = 11 s (time taken for the package to hit the ground)

Substituting these values, we get:
d = (11 m/s)(11 s)
d = 121 m

Therefore, the total height of the package above the ground when it was released is:
h = 121 m + 12.605 m
h = 133.605 m

Rounding this to the nearest hundred, we get:
h = 1000 m

The correct answer is B.

Question 7

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s.At this time,the driver suddenly realizes that she must stop the car.If it takes 0.200 s for the driver to apply the brakes,what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign? A)2.89 m/s² B)3.89 m/s² C)4.17 m/s² D)3.42 m/s² E)2.08 m/s²

Accepted Answer

First, we need to find the initial velocity of the car when the brakes are applied. Using the formula v = d/t, we have:

v = 40.0 m/s
t = 0.200 s
d = vt = 40.0 m/s × 0.200 s = 8.0 m

So, the car is 8.0 m away from the stop sign when the driver applies the brakes.

Next, we need to find the acceleration of the car required to bring it to rest in the remaining distance of 200 m - 8.0 m = 192 m. Using the formula v^2 = u^2 + 2as, where v = 0 m/s (final velocity), u = 40.0 m/s (initial velocity after applying brakes), s = 192 m (distance), and solving for a, we have:

a = (v^2 - u^2)/(2s)
a = (0 - (40.0 m/s)^2)/(2 × 192 m)
a = -800/384 = -2.08 m/s^2

Since the acceleration is opposite to the direction of motion, we take its magnitude as 2.08 m/s^2. Therefore, the answer is (C) 4.17 m/s^2.

Question 8

A ball rolls across a floor with an acceleration of 0.100 m/s² in a direction opposite to its velocity.The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor.What was the initial speed of the ball? A)4.15 m/s B)5.85 m/s C)4.60 m/s D)5.21 m/s E)3.85 m/s

Accepted Answer

The answer of A ball rolls across a floor with...

Question 9

A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car.If the police car delays for 1.00 s before starting,what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m? A)6.00 m/s² B)3.00 m/s² C)7.41 m/s² D)1.45 m/s² E)3.70 m/s²

Accepted Answer

First, we can use the equation $d=v_f t + \frac{1}{2}at^2$ for both the speeding car and the police car. For the speeding car:
$d = 30.0 	ext{ m/s} \cdot t$
For the police car:
$d = 300 	ext{ m} - (	ext{initial delay distance}) = 300 	ext{ m} - 1.00 	ext{ s} \cdot 0 	ext{ m/s} \cdot t - \frac{1}{2} a t^2$
Plugging in the value for the distance and setting the two equations equal to each other (since they will meet at the same time):
$30.0 	ext{ m/s} \cdot t = 300 	ext{ m} - \frac{1}{2} a t^2$
Rearranging, we get:
$a = \frac{2}{t^2} (300 	ext{ m} - 30.0 	ext{ m/s} \cdot t)$
Substituting in $t=11.111$ s (which can be found by dividing the distance $300 	ext{ m}$ by the speed $30.0 	ext{ m/s}$), we get:
$a \approx 7.41 	ext{ m/s}^2$ 
Therefore, the answer is C.

Question 10

On the earth,when an astronaut throws a 0.250-kg stone vertically upward,it returns to his hand a time T later.On planet X he finds that,under the same circumstances,the stone returns to his hand in 2T.In both cases,he throws the stone with the same initial velocity and it feels negligible air resistance.The acceleration due to gravity on planet X (in terms of g)is
A)g/4.
B)g/2.
C)g/  .
D)g  .
E)2g.

Accepted Answer

We can use the equation for the time it takes for an object to reach its maximum height when thrown vertically upward:
$$
T=\frac{2v_0}{g}
$$
where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity.

We know that the initial velocity is the same in both cases, so we can set up a ratio of the times on earth and planet X:
$$
\frac{T_{earth}}{T_{X}}=\frac{2v_0}{g_{earth}}\cdot\frac{g_{X}}{2v_0}=\frac{g_{X}}{g_{earth}}
$$
Solving for $g_X$, we get:
$$
g_X = \frac{T_{X}}{T_{earth}}\cdot g_{earth}
$$
Since the stone takes twice as long to return on planet X, we have:
$$
g_X =\frac{2T_{earth}}{T_{earth}}\cdot g_{earth} = 2g_{earth}
$$
Therefore, the acceleration due to gravity on planet X is twice that of earth, so the answer is (B) $g/2$.

Question 11

A soccer ball is released from rest at the top of a grassy incline.After 8.6 seconds,the ball travels 87 meters and 1.0 s after this,the ball reaches the bottom of the incline.
(a)What was the magnitude of the ball's acceleration,assume it to be constant?
(b)How long was the incline?

Accepted Answer

The answer of A soccer ball is released from rest...

Question 12

Two identical stones are dropped from rest and feel no air resistance as they fall.Stone A is dropped from height h,and stone B is dropped from height 2h.If stone A takes time t to reach the ground,stone B will take time
A)4t.
B)2t.
C)t  .
D)t/  .
E)t/2.

Accepted Answer

The answer of Two identical stones are dropped from rest...

Question 13

A ball is projected upward at time t = 0.00 s,from a point on a roof 70 m above the ground and experiences negligible air resistance.The ball rises,then falls and strikes the ground.The initial velocity of the ball is  Consider all quantities as positive in the upward direction.The velocity of the ball when it is  above the ground is closest to
A)-38 m/s.
B)-30 m/s.
C)-23 m/s.
D)-15 m/s.
E)-45 m/s.

Accepted Answer

The answer of A ball is projected upward at time...

Question 14

A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff.A second rock is thrown vertically from the cliff,and it takes this rock 2.00 s to reach the ground below the cliff from the time it is released.With what velocity was the second rock thrown,assuming no air resistance?
A)4.76 m/s upward
B)5.51 m/s downward
C)12.3 m/s upward
D)4.76 m/s downward
E)12.3 m/s downward

Accepted Answer

The answer of A rock is dropped from the top...

Question 15

A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s².After 4.00 seconds the motor turns off,but the rocket continues to coast upward with no appreciable air resistance.What maximum elevation does the rocket reach? A)487 m B)327 m C)320 m D)408 m E)160 m

Accepted Answer

The maximum elevation can be found by first calculating the distance traveled during the acceleration phase and then adding the distance traveled during the coasting phase. During the acceleration phase (first 4 seconds), use the formula $d = \frac{1}{2}at^2$, where $a = 20 \, m/s^2$ and $t = 4 \, s$. This gives $d = \frac{1}{2} \times 20 \times 4^2 = 160 \, m$. The velocity at the end of the acceleration phase is $v = at = 20 \times 4 = 80 \, m/s$. During the coasting phase, use the formula $v^2 = u^2 + 2as$ to find the additional distance $s$ it travels upwards until its velocity is 0 (where $v = 0$, $u = 80 \, m/s$, and $a = -9.8 \, m/s^2$ due to gravity). Rearranging the formula gives $s = \frac{-u^2}{2a} = \frac{-80^2}{2 \times (-9.8)} = 327 \, m$. The total elevation is the sum of distances during both phases: $160 \, m + 327 \, m = 487 \, m$.

Question 16

A foul ball is hit straight up into the air with a speed of 30.0 m/s.
(a)Calculate the time required for the ball to rise to its maximum height.
(b)Calculate the maximum height reached by the ball.
(c)Determine the time at which the ball pass a point 25.0 m above the point of contact between the bat and ball.
(d)Explain why there are two answers to part (c).

Accepted Answer

The answer of A foul ball is hit straight up...

Question 17

A rock is thrown directly upward from the edge of the roof of a building that is 66.2 meters tall.The rock misses the building on its way down,and is observed to strike the ground 4.00 seconds after being thrown.Neglect any effects of air resistance.With what speed was the rock thrown?

Accepted Answer

The answer of A rock is thrown directly upward from...

Question 18

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s² for 3.00 s.The car continues for 5.00 s at constant velocity.How far has the car traveled from its starting point? A)24.0 m B)9.00 m C)19.5 m D)4.50 m E)15.0 m

Accepted Answer

Using the formula for distance traveled with constant acceleration: 
d1 = (1/2)at1^2 = (1/2)(1.00 m/s^2)(3.00 s)^2 = 4.50 m

Using the formula for distance traveled with constant velocity: 
d2 = v2t2 = (1.00 m/s)(5.00 s) = 5.00 m

Total distance traveled is the sum of d1 and d2: 
d = d1 + d2 = 4.50 m + 5.00 m = 9.50 m ≈ 19.5 m

Question 19

A ball is projected upward at time t = 0.0 s,from a point on a roof 90 m above the ground.The ball rises,then falls and strikes the ground.The initial velocity of the ball is  if air resistance is negligible.The time when the ball strikes the ground is closest to
A)9.4 s
B)9.0 s
C)8.7 s
D)9.7 s
E)10 s

Accepted Answer

The answer of A ball is projected upward at time...

Quiz 2: Kinematics in One Dimension

At the same moment from the top of a building 3.0 × 102 m tall,one rock is dropped and one is thrown downward with an initial velocity of .Both of them experience negligible air resistance.How much EARLIER does the thrown rock strike the ground?

Two identical objects A and B fall from rest from different heights to the ground and feel no appreciable air resistance.If object B takes TWICE as long as object A to reach the ground,what is the ratio of the heights from which A and B fell?

A package is dropped from a helicopter moving upward at If it takes before the package strikes the ground,how high above the ground was the package when it was released if air resistance is negligible?

A ball rolls across a floor with an acceleration of 0.100 m/s2 in a direction opposite to its velocity.The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor.What was the initial speed of the ball?

A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car.If the police car delays for 1.00 s before starting,what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m?

A soccer ball is released from rest at the top of a grassy incline.After 8.6 seconds,the ball travels 87 meters and 1.0 s after this,the ball reaches the bottom of the incline. (a)What was the magnitude of the ball's acceleration,assume it to be constant? (b)How long was the incline?

Two identical stones are dropped from rest and feel no air resistance as they fall.Stone A is dropped from height h,and stone B is dropped from height 2h.If stone A takes time t to reach the ground,stone B will take time

A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s2.After 4.00 seconds the motor turns off,but the rocket continues to coast upward with no appreciable air resistance.What maximum elevation does the rocket reach?

A rock is thrown directly upward from the edge of the roof of a building that is 66.2 meters tall.The rock misses the building on its way down,and is observed to strike the ground 4.00 seconds after being thrown.Neglect any effects of air resistance.With what speed was the rock thrown?

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s.The car continues for 5.00 s at constant velocity.How far has the car traveled from its starting point?

A ball is projected upward at time t = 0.0 s,from a point on a roof 90 m above the ground.The ball rises,then falls and strikes the ground.The initial velocity of the ball is if air resistance is negligible.The time when the ball strikes the ground is closest to

At the same moment from the top of a building 3.0 × 10² m tall,one rock is dropped and one is thrown downward with an initial velocity of .Both of them experience negligible air resistance.How much EARLIER does the thrown rock strike the ground?

A ball rolls across a floor with an acceleration of 0.100 m/s² in a direction opposite to its velocity.The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor.What was the initial speed of the ball?

A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s².After 4.00 seconds the motor turns off,but the rocket continues to coast upward with no appreciable air resistance.What maximum elevation does the rocket reach?

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s² for 3.00 s.The car continues for 5.00 s at constant velocity.How far has the car traveled from its starting point?