Question 1

The figure represents the parabolic trajectory of a ball going from a to e in Earth gravity but without air resistance. The initial velocity of the ball is V_o at an angle $\theta$ to the horizon. The vertical dashed lines represent equal time interval, $\Delta$t. The acceleration at point a is ____ the acceleration at c. A) less than B) equal to C) greater than D) Unable to tell. E) Depends on how g changes along the path.

Accepted Answer

The acceleration of the ball is constant (equal to g, the acceleration due to gravity) throughout its trajectory in the absence of air resistance. Therefore, the acceleration at point a is equal to the acceleration at point c.

Question 2

Which of the following statements is not true of a projectile moving against negligible air resistance near the surface of Earth?
A) The horizontal velocity is constant.
B) The vertical acceleration is constant.
C) The horizontal displacement is directly proportional to the time of flight.
D) The vertical velocity at any given time is independent of the angle of projection.
E) The horizontal acceleration is constant.

Accepted Answer

The vertical velocity at any given time is not independent of the angle of projection. The vertical velocity is a combination of the initial velocity and the acceleration due to gravity, which is dependent on the angle of projection. The greater the angle of projection, the greater the initial vertical velocity and the greater the vertical acceleration due to gravity.

Question 3

The maximum horizontal range of a rock that is thrown at the same speed but different angles with the horizon and that lands at a level, H, below the initial level occurs when the angle is
A) less than 0$\degree$.
B) 0$\degree$.
C) greater than 0$\degree$ and less than 45$\degree$ actual value depends on the value of H).
D) 45$\degree$,
E) greater than 45$\degree$.

Accepted Answer

The maximum range occurs when the projectile is launched at a 45-degree angle with respect to the horizontal. However, the question specifies that the rock must land at a level below the initial level. This means that the initial height above the ground affects the optimal angle. As the initial height decreases, the optimal angle becomes smaller than 45 degrees. Therefore, the correct answer is "greater than 0 degrees and less than 45 degrees (actual value depends on the value of H)."

Question 4

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance.  At point C the velocity of the ball is
A) a maximum and is directed to the right.
B) directed to the left.
C) a maximum.
D) a minimum and is directed to the right.
E) zero.

Accepted Answer

The answer of The figure represents the parabolic trajectory of...

Question 5

The figure represents the parabolic trajectory of a ball going from a to e in Earth gravity but without air resistance. The initial velocity of the ball is V_o at an angle $\theta$ to the horizon. The vertical dashed lines represent equal time interval, $\Delta$t. Consider "up" as the positive direction. The vertical velocity at point b is [Note: g = 9.81 m/s²)] A) 0. B) V_o cos $\theta$) - ½ g 2$\Delta$t)². C) V_o sin $\theta$) - ½ g 2$\Delta$t)². D) V_o sin$\theta$) + ½ g 2$\Delta$t)². E) V_o cos $\theta$) + ½ g 2$\Delta$t)².

Accepted Answer

At point b, the ball has reached its highest point in the trajectory, so the vertical velocity is zero. From the conservation of energy, we know that the initial total energy equals the final total energy. At point a, the initial energy is given by: Ea = ½ m V_o² At point b, the potential energy is converted to kinetic energy and we have: Eb = ½ m Vb² where Vb is the velocity magnitude at point b. Therefore, we have: Ea = Eb ½ m V_o² = ½ m Vb² + mgh where h is the height of point b relative to point a. Since h is maximum at point b, we can write: h = V_o² sin² F1F1F1S1F1F1F10) / 2g Substituting h into the previous equation, we get: ½ m V_o² = ½ m Vb² + V_o² sin² F1F1F1S1F1F1F10) / 2 Solving for Vb, we obtain: Vb = V_o sin F1F1F1S1F1F1F10) - g sin F1F1F1S1F1F1F10) F1F1F1S1F1F1F10t / 2 The negative sign in front of the g term indicates that the velocity in the vertical direction is decreasing due to the Earth's gravity. Therefore, the correct answer is option C: V_o sin F1F1F1S1F1F1F10) - ½ g 2F1F1F1S1F1F1F10t)².

Question 6

A plane is flying horizontally at a height of 500 m and a constant speed of 429 km/h when an object is projected vertically downward at an initial speed of 35.0 m/s. If air resistance is neglected, the average vertical component of velocity between the object's release and its striking the ground is A) 70.0 m/s. B) 5.50 * 10³ m/s. C) 140 m/s. D) 105 m/s. E) 35.0 m/s.

Accepted Answer

The average vertical component of velocity can be calculated using the formula for the final velocity of an object in free fall, $v = u + gt$, where $v$ is the final velocity, $u$ is the initial velocity (35.0 m/s downward), $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$ downward), and $t$ is the time of fall. Since the object is projected downward, both $u$ and $g$ act in the same direction, increasing the object's speed as it falls.To find the average velocity, we first need to determine the time of fall ($t$). This can be found using the equation $s = ut + \frac{1}{2}gt^2$, where $s$ is the distance fallen (500 m). However, to find the average velocity directly without finding $t$, we can use the fact that the average velocity ($v_{avg}$) in uniformly accelerated motion is $\frac{u + v}{2}$, where $v$ is the final velocity.Given that we don't have the final velocity directly, and to simplify the explanation, let's correct the approach by focusing on the concept that the average velocity in this context is essentially the mean of the initial and final velocities in the vertical direction. The final velocity can be calculated using the kinematic equations, but the question seems to be aimed at understanding the concept rather than the detailed calculation.Given the initial downward speed and the acceleration due to gravity, the object will accelerate downwards, increasing its downward velocity over the time it takes to hit the ground. The average vertical component of the velocity would be higher than the initial speed due to this acceleration.However, without the exact time of fall or final velocity calculated, we cannot directly apply the formulas as intended in this simplified explanation. The correct approach involves calculating the time of fall using the given height and initial vertical speed, then finding the final velocity using $v = u + gt$, and finally calculating the average of the initial and final velocities. But, given the options and the nature of the question, it seems to expect an understanding that the average velocity will be significantly affected by the acceleration due to gravity, leading to an average that is higher than the initial velocity but not explicitly calculable with the provided information.Therefore, without the detailed calculations based on the provided options and aiming for a simplified explanation, the correct choice would involve recognizing that the average vertical velocity is affected by gravity and would be higher than the initial projection speed. However, the options provided (like "B" with unclear symbols) and the lack of direct calculation may lead to confusion.For a precise answer, detailed calculations involving the time of fall and the final velocity would be necessary, which are not fully addressed in this simplified explanation. The aim was to clarify the concept rather than provide a misleading calculation.

Question 7

The figure represents the parabolic trajectory of a ball going from a to e in Earth gravity but without air resistance. The initial velocity of the ball is V_o at an angle $\theta$ to the horizon. The vertical dashed lines represent equal time interval, $\Delta$t. Consider UP as the positive direction. The vertical velocity at point c is [Note: g = 9.81 m/s²)] A) V_o cos $\theta$). B) V_o cos $\theta$) - ½ g 3$\Delta$t)². C) V_o sin $\theta$) - ½ g 3$\Delta$t)². D) 0. E) V_o cos $\theta$) + ½ g 3$\Delta$t)².

Accepted Answer

The answer of The figure represents the parabolic trajectory of...

Question 8

A projectile was fired at 35º above the horizontal. At the highest point in its trajectory its speed was 200 m/s. If air resistance is ignored, the initial velocity had a horizontal component of
A) zero.
B) 200 cos 35º) m/s.
C) 200 sin 35º) m/s.
D) 200/cos 35º) m/s.
E) 200 m/s.

Accepted Answer

The answer of A projectile was fired at 35º above...

Question 9

A projectile is fired horizontally from the top of a building in Earth gravity with an initial speed v₀. In the absence of air resistance, which of the graphs is representative of the projectile's motion? A) 1) B) 2) C) 3) D) 4) E) 5)

Accepted Answer

The projectile fired horizontally from the top of a building will undergo projectile motion in which it will follow a parabolic path. The vertical component of its velocity will be affected by gravity and will cause the projectile to accelerate downwards with a constant acceleration of 9.8 m/s^2. However, the horizontal component of its velocity will remain constant throughout its motion. Thus, the graph that best represents this motion is graph 1), which shows a parabolic path indicating a constant horizontal velocity and a downward acceleration due to gravity.

Question 10

If a baseball is thrown at an angle of elevation of 30°, its range in the absence of air resistance is less than that of one thrown at
A) 0º.
B) 80º.
C) 20º.
D) 55º.
E) 90º.

Accepted Answer

The answer of If a baseball is thrown at an...

Question 11

The figure represents the parabolic trajectory of a ball going from a to e in Earth gravity but without air resistance. The initial velocity of the ball is V_o at an angle $\theta$ to the horizon. The vertical dashed lines represent equal time interval, $\Delta$t. The horizontal velocity at point c is A) 0. B) V_o cos $\theta$). C) V_o sin $\theta$). D) V_o. E) depends on what g is along the path.

Accepted Answer

At point c, the ball has reached its maximum height and is momentarily at rest before falling back down. Therefore, its vertical velocity is 0 at this point. Since there is no air resistance, the horizontal velocity remains constant throughout the entire trajectory. Thus, the horizontal velocity at point c is equal to the initial horizontal velocity, which is V_o cos F1F1F1S1F1F1F10).

Question 12

If you ignore air resistance in projectile motion, the ________ of the projectile remains constant.
A) velocity vector
B) horizontal component of the velocity vector
C) vertical component of the velocity vector
D) speed
E) range

Accepted Answer

Ignoring air resistance, the horizontal component of the velocity remains constant because there is no horizontal force acting on the projectile. Gravity only affects the vertical component, causing it to change.

Question 13

A golfer drives her ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point of its flight,
A) its velocity and acceleration are both zero.
B) its velocity is zero but its acceleration is nonzero.
C) its velocity is nonzero but its acceleration is zero.
D) its velocity and acceleration are both nonzero.
E) Insufficient information is given to answer correctly.

Accepted Answer

The answer of A golfer drives her ball from the...

Question 14

A projectile is fired with an initial speed of 1000 m/s at an angle of 37° above the horizontal. If air resistance is neglected, the horizontal component of the projectile's velocity after 20 s is approximately
A) 600 m/s.
B) 800 m/s.
C) 640 m/s.
D) 40 m/s,
E) 160 m/s.

Accepted Answer

The answer of A projectile is fired with an initial...

Question 15

A baseball is thrown with a velocity at an angle $\theta$ with the horizontal. The horizontal component of its velocity in the absence of air resistance is A) v cos $\theta$ B) v cos $\theta$ - gt. C) v² + 2gx. D) v² - 2gx. E) v sin $\theta$ - gt.

Accepted Answer

The horizontal component of velocity is given by v cosθ, where θ is the angle of the velocity vector with the horizontal. In this case, the angle is given as θ= F1F1F1S1θF1F1F10. Therefore, the horizontal component of velocity is v cos( F1F1F1S1θF1F1F10) which corresponds to choice A.

Question 16

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance.  What is the direction of the acceleration at point C?
A) It is to the right.
B) It is to the left.
C) It is straight up.
D) It is straight down.
E) The acceleration of the ball is zero.

Accepted Answer

At point C, the ball reaches its maximum height and starts to fall back down, so the direction of acceleration must be straight down, opposing the velocity of the ball.

Question 17

You shoot an arrow in a high arc toward a target located some distance away. At the highest point in its flight, the arrow's
A) velocity and acceleration are both nonzero.
B) velocity is zero but its acceleration is nonzero.
C) velocity is nonzero but its acceleration is zero.
D) velocity and acceleration are both zero.
E) Insufficient information is given to answer correctly.

Accepted Answer

At the highest point in its flight, the arrow's velocity is not zero; it has a horizontal component of velocity. However, its vertical component of velocity is zero. The acceleration due to gravity is still acting on the arrow, giving it a nonzero acceleration downwards.

Question 18

A projectile is shot at an angle of 45º to the horizontal near the surface of Earth but in the absence of air resistance. When it reaches the highest point of its trajectory, its speed is 150 m/s. In a second trial with the same projectile, the initial speed is the same but the angle is now 37º with the horizontal. At its highest point in this trajectory, the speed of the projectile would be
A) 150 m/s sin 45º/sin 36º).
B) 150 m/s cos 37º/cos 45º).
C) 150 m/s sin 37º/sin 45º).
D) 150 m/s (37/45).
E) None of these is correct.

Accepted Answer

At the highest point in its trajectory, the vertical component of the projectile's velocity is zero, and only the horizontal component remains. The horizontal component of velocity remains constant throughout the flight in the absence of air resistance. The speed at the highest point is given by the initial speed multiplied by the cosine of the launch angle (since cos(θ) gives the horizontal component of the initial velocity vector). Therefore, for the second trial, the speed at the highest point is 150 m/s multiplied by the ratio of the cosines of the new and original launch angles, which is cos(37º)/cos(45º).

Question 19

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance.  What is the direction of the acceleration at point B?
A) It is up and to the right.
B) It is down and to the left.
C) It is straight up.
D) It is straight down.
E) The acceleration of the ball is zero.

Accepted Answer

At point B, the ball has reached the highest point of its trajectory and is starting to fall back towards the ground. Therefore, the acceleration of the ball must be straight down towards the ground, which is represented by answer choice D.

Question 20

A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when it is at an altitude of 600 m. If air resistance is ignored, the horizontal distance of the point of impact from the plane at the moment of the package's release is A) 2.80 * 10³ m. B) 720 m. C) 6.80 * 10³ m. D) 420 m. E) 5.50 * 10³ m.

Accepted Answer

To find the horizontal distance, we first need to calculate the time it takes for the package to hit the ground. Ignoring air resistance, the vertical component of the velocity and the altitude can be used to calculate this time. The vertical velocity component (Vv) can be found using the sine of the angle (sin(37º)) and the total velocity (250 m/s), so Vv = 250 * sin(37º). However, the exact calculation of time involves the formula derived from the equation of motion under gravity, which is not directly shown here. Once the time (t) is found, the horizontal distance (d) can be calculated using the horizontal component of the velocity (Vh = 250 * cos(37º)) multiplied by time (t). The correct answer does not require the explicit calculation of these components but understanding the process leads to the realization that option B (720 m) is the correct choice based on standard projectile motion equations and given options.

Quiz 3: Motion in Two and Three Dimensions

Which of the following statements is not true of a projectile moving against negligible air resistance near the surface of Earth?

The maximum horizontal range of a rock that is thrown at the same speed but different angles with the horizon and that lands at a level, H, below the initial level occurs when the angle is

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance. At point C the velocity of the ball is

A projectile was fired at 35º above the horizontal. At the highest point in its trajectory its speed was 200 m/s. If air resistance is ignored, the initial velocity had a horizontal component of

A projectile is fired horizontally from the top of a building in Earth gravity with an initial speed v0. In the absence of air resistance, which of the graphs is representative of the projectile's motion?

If a baseball is thrown at an angle of elevation of 30°, its range in the absence of air resistance is less than that of one thrown at

If you ignore air resistance in projectile motion, the ________ of the projectile remains constant.

A golfer drives her ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point of its flight,

A projectile is fired with an initial speed of 1000 m/s at an angle of 37° above the horizontal. If air resistance is neglected, the horizontal component of the projectile's velocity after 20 s is approximately

A baseball is thrown with a velocity at an angle θ\thetaθ with the horizontal. The horizontal component of its velocity in the absence of air resistance is

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance. What is the direction of the acceleration at point C?

You shoot an arrow in a high arc toward a target located some distance away. At the highest point in its flight, the arrow's

The figure represents the parabolic trajectory of a ball going from A to E in Earth gravity but without air resistance. What is the direction of the acceleration at point B?

A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when it is at an altitude of 600 m. If air resistance is ignored, the horizontal distance of the point of impact from the plane at the moment of the package's release is

A projectile is fired horizontally from the top of a building in Earth gravity with an initial speed v₀. In the absence of air resistance, which of the graphs is representative of the projectile's motion?