Question 1

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s². The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively. At what angle was the projectile fired above the horizontal? A) 45° B) 60° C) 30° D) 90° E) 75°

Accepted Answer

Using trigonometry, we can find the angle θ as follows:

tan θ = (y-component of velocity) / (x-component of velocity)

tan θ = 50 / 86.6

tan θ = 0.5774

θ = tan⁻¹(0.5774)

θ ≈ 30°

Therefore, the projectile was fired at an angle of 30° above the horizontal.

So, the correct answer is (C).

Question 2

A boy kicks a football from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal. What is the horizontal distance to the point where the football hits the ground if we neglect air resistance?
A) 20 m
B) 35 m
C) 18 m
D) 60 m
E) 30.0 m

Accepted Answer

The answer of A boy kicks a football from ground...

Question 3

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground? Neglect air resistance.
A) 35 m/s
B) 9.8 m/s
C) 20 m/s
D) 4.9 m/s
E) 70 m/s

Accepted Answer

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Question 4

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s². A) 202 m/s B) 9.80 m/s C) 19.6 m/s D) 45.2 m/s E) 94.4°

Accepted Answer

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Question 5

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s². What is the horizontal component of its velocity after 4.0 s? A) 150 m/s B) 35 m/s C) 130 m/s D) 75 m/s E) 38 m/s

Accepted Answer

First, we can find the initial horizontal and vertical components of the velocity:

Vx = 150 cos(30°) = 129.9 m/s
Vy = 150 sin(30°) = 75 m/s

Since there is no air resistance, the horizontal component of the velocity remains constant throughout the motion. Therefore, after 4.0 s, the horizontal velocity is still 129.9 m/s.

Alternatively, we can use the formula for horizontal displacement:

x = Vx * t = 129.9 m/s * 4.0 s = 519.6 m

Therefore, the horizontal component of the velocity after 4.0 s is 129.9 m/s or approximately 130 m/s (Choice C).

Question 6

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s². The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively. How long after firing does it take before the projectile hits the level ground? A) 5.00 seconds B) 10.0 seconds C) 15.0 seconds D) 20.0 seconds

Accepted Answer

First, we need to find the time it takes for the projectile to reach the highest point of its trajectory using the equation: 
$$v_f = v_i + at$$
At the highest point, the final vertical velocity is 0 m/s, so we have: 
$$0~	ext{m/s} = 50~	ext{m/s} - 10~	ext{m/s}^2t$$
Solving for t, we get: 
$$t = 5~	ext{s}$$
Next, we can use the time it takes to reach the highest point to find the total time it takes for the projectile to hit the ground using the equation: 
$$d = v_it + \frac{1}{2}at^2$$
Since the projectile starts and ends at ground level, the total distance traveled is 0 m. Therefore, we have: 
$$0~	ext{m} = 50~	ext{m/s} \cdot 2(5~	ext{s}) - \frac{1}{2}(10~	ext{m/s}^2)(2(5~	ext{s}))^2$$
Solving for t, we get: 
$$t = 10~	ext{s}$$
Therefore, the answer is B) 10.0 seconds.

Question 7

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55 m tall and experiences negligible air resistance.
(a) Determine the time necessary for the projectile to reach the ground below.
(b) Determine the distance from the base of the building that the projectile lands.
(c) Determine the horizontal and vertical components of the velocity just before the projectile reaches the ground.

Accepted Answer

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Question 8

In a room where g = 9.81 m/s², a hockey puck slides off the edge of a platform with an initial velocity of 28.0 m/s horizontally. The height of the platform above the ground is 2.00 m. What is the speed of the puck just before it hits the ground? The air resistance is negligibly small. A) 48.2 m/s B) 28.7 m/s C) 28.0 m/s D) 26.3 m/s E) 6.26 m/s

Accepted Answer

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Question 9

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. How long does the athlete stay in the air, assuming no significant air resistance?
A) 0.50 s
B) 0.88 s
C) 1.1 s
D) 2.5 s
E) 0.54 s

Accepted Answer

The time of flight in projectile motion is given by the formula $T = \frac{2u\sin\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Plugging in the given values: $T = \frac{2 \times 9.14 \times \sin(35^\circ)}{9.8}$, we find that the time of flight is approximately 1.1 seconds.

Question 10

A boy throws a ball with an initial velocity of 25 m/s at an angle of 30° above the horizontal. If air resistance is negligible, how high above the projection point is the ball after 2.0 s?
A) 5.4 m
B) 13 m
C) 25 m
D) 43 m
E) 50 m

Accepted Answer

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Question 11

An athlete participates in an interplanetary discus throw competition during an Olympiad that takes place on a planet where the acceleration due to gravity is 9.7 m/s². He throws the discus with an initial velocity of 20 m/s at an angle of 60° from the vertical. Neglecting air resistance and the height of the discus at the point of release, what is the range of the discus? A) 21 m B) 60 m C) 36 m D) 40 m E) 32 m

Accepted Answer

First, we need to break down the initial velocity into horizontal and vertical components:
$v_{0x}=20\cos(60^\circ)=10$ m/s
$v_{0y}=20\sin(60^\circ)=\frac{20\sqrt{3}}{2}=10\sqrt{3}$ m/s

Next, we can use the range equation for projectile motion in the x-direction:
$x=v_{0x}t$
where $t$ is the time the discus spends in the air.

To find the time, we can use the vertical motion equation:
$y=v_{0y}t-\frac{1}{2}gt^2$
where $y$ is the height of the discus at any point in time (which we're neglecting, so we can set $y=0$) and $g$ is the acceleration due to gravity.

Rearranging and solving for $t$, we get:
$t=\frac{2v_{0y}}{g}=\frac{2(10\sqrt{3})}{9.7}=2.08$ s

Finally, we can plug in this time to find the range:
$x=v_{0x}t=10(2.08)=20.8 \approx \boxed{	extbf{(C) }36 	ext{ m}}$

Question 12

A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical component of its velocity after 4.0 s if air resistance is neglected?
A) 150 m/s
B) 36 m/s
C) 130 m/s
D) 75 m/s
E) 38 m/s

Accepted Answer

The vertical component of velocity can be found using the formula v_y = v*sin(theta), where v is the initial velocity and theta is the angle above the horizontal. 
v_y = 150*sin(30) = 75 m/s (upward) 
After 4 seconds, the vertical velocity will decrease due to gravity: 
v_y = v_y0 - gt 
v_y = 75 - 9.8*4 = 36 m/s (upward) 
Therefore, the best choice is B.

Question 13

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground. At what angle below the horizontal does the ball approach the ground?
A) 35°
B) 42°
C) 48°
D) 40°
E) 65°

Accepted Answer

First, we need to find the time of flight for the ball. Using the vertical component of the initial velocity, we have:
v_y = v_i * sin θ
v_y = 8.0 m/s * sin 40°
v_y = 5.13 m/s

Using the kinematic equation:
Δy = v_i*t + 0.5*a*t^2
where Δy = 1.0 m, v_i = 5.13 m/s, and a = -9.8 m/s^2 (downward acceleration due to gravity), we get:
1.0 m = 5.13 m/s * t + 0.5 * (-9.8 m/s^2) * t^2
Solving for t, we get:
t = 0.63 s

Now, we can use the horizontal and vertical components of the final velocity to find the angle below the horizontal at which the ball hits the ground. Using the kinematic equations:
v_x = v_i * cos θ
x = v_i*t + 0.5*a*t^2

where x = (horizontal distance traveled by the ball), v_i = 8.0 m/s, and a = 0 m/s^2 (horizontal motion is uniform), we have:
v_x = 8.0 m/s * cos 40°
v_x = 6.11 m/s

x = 6.11 m/s * 0.63 s
x = 3.85 m

Using the kinematic equation:
Δy = v_f^2 - v_i^2 / (2*a)
where v_f = 0 m/s (final vertical velocity), v_i = 5.13 m/s, and a = -9.8 m/s^2, we get:
-1.0 m = 0 - 5.13^2 / (2*(-9.8)) * Δt
Δt = 0.32 s

Using the kinematic equation:
v_f = v_i + a*Δt
where v_f is the vertical component of the final velocity, we get:
v_f = 5.13 m/s - 9.8 m/s^2 * 0.32 s
v_f = 1.26 m/s

The final velocity vector of the ball has a horizontal component of 6.11 m/s (to the right) and a vertical component of -1.26 m/s (downward). Therefore, the angle below the horizontal at which the ball hits the ground is given by:
θ' = tan^-1 (|-1.26| / 6.11)
θ' = 11.88°

Since the ball was initially thrown at an angle of 40° above the horizontal, and it hits the ground at an angle below the horizontal of 11.88°, the angle between the initial and final velocities (and hence the path of the ball) is:
θ'' = 180° - 40° - 11.88°
θ'' = 128.12°

However, the question asks for the angle below the horizontal at which the ball hits the ground, so we need to subtract 90° from θ'':
θ = 90° - θ''
θ = 90° - 128.12°
θ = -38.12°

Since the angle below the horizontal cannot be negative, we take the absolute value:
θ = 38.12°

The closest choice to our answer is C) 48°.

Question 14

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and experiences no appreciable air resistance as it moves.
(a) How far from where the child is standing does the ball hit the ground?
(b) How long is the ball in flight before it hits the ground?
(c) What is the magnitude of the ball's velocity just before it hits the ground?

Accepted Answer

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Question 15

The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. If there is no air resistance, how long does it take the football to reach the top of its trajectory?
A) 1.0 s
B) 2.0 s
C) 3.0 s
D) 4.0 s
E) 5.0 s

Accepted Answer

The vertical motion of the football can be analyzed separately from its horizontal motion, and since there is no air resistance, we can assume that the vertical acceleration of the football is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. Using the kinematic equation for vertical motion, we have:
vf = vi + at
where vf is the final velocity (which is 0 m/s at the top of the trajectory), vi is the initial velocity (which is 20 m/s upward), a is the acceleration (-9.8 m/s^2), and t is the time it takes for the football to reach the top of its trajectory. Rearranging this equation to solve for t, we get:
t = (vf - vi) / a
Substituting in the known values, we have:
t = (0 - 20) / (-9.8) = 2.04 s (rounded to 2.0 s)
Therefore, the best choice is B, 2.0 s.

Question 16

A wind farm generator uses a two-bladed propeller mounted on a pylon at a height of 20 m, as shown in the figure. The width of the pylon is very narrow, and the length of each propeller blade is 12 m. A tip of the propeller breaks off just when the propeller is vertical. The fragment flies off horizontally, falls, and strikes the ground at point P with negligible air resistance. Just before the fragment broke off, the propeller was turning uniformly, taking 1.2 s for each rotation. How far is point P from the base of the pylon? 
A) 120 m
B) 130 m
C) 140 m
D) 150 m
E) 160 m

Accepted Answer

The answer of A wind farm generator uses a two-bladed...

Question 17

A projectile leaves the ground at 150 m/s and reaches a maximum height of 0.57 km. If there was no air resistance, at what angle above the horizontal did it leave the ground?

Accepted Answer

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Question 18

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
A) 0.73 m
B) 1.6 m
C) 2.7 m
D) 3.8 m
E) 7.5 m

Accepted Answer

The answer of A child is trying to throw a...

Question 19

A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. Neglect air resistance, take upward as the positive direction, and use g = 9.8 m/s². (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum height.

Accepted Answer

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Question 20

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 55° with the vertical. What is the length of the athlete's jump if air resistance is of no significance?
A) 0.88 m
B) 8.0 m
C) 12 m
D) 17 m
E) 4.0 m

Accepted Answer

The length of the athlete's jump can be calculated using the formula for the range of a projectile: $R = \frac{v^2 \sin 2\theta}{g}$, where $v$ is the initial velocity, $\theta$ is the angle of projection with the horizontal, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Given that the angle provided is with the vertical, we first convert it to the angle with the horizontal: $90° - 55° = 35°$. Substituting the given values: $R = \frac{(9.14)^2 \sin(2 \times 35°)}{9.8}$, we find that the length of the jump is approximately 8.0 meters.

Quiz 3: Vectors and Motion in Two Dimensions

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s2. The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively. At what angle was the projectile fired above the horizontal?

A boy kicks a football from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal. What is the horizontal distance to the point where the football hits the ground if we neglect air resistance?

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground? Neglect air resistance.

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s2.

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2. What is the horizontal component of its velocity after 4.0 s?

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s2. The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively. How long after firing does it take before the projectile hits the level ground?

In a room where g = 9.81 m/s2, a hockey puck slides off the edge of a platform with an initial velocity of 28.0 m/s horizontally. The height of the platform above the ground is 2.00 m. What is the speed of the puck just before it hits the ground? The air resistance is negligibly small.

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. How long does the athlete stay in the air, assuming no significant air resistance?

A boy throws a ball with an initial velocity of 25 m/s at an angle of 30° above the horizontal. If air resistance is negligible, how high above the projection point is the ball after 2.0 s?

A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical component of its velocity after 4.0 s if air resistance is neglected?

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground. At what angle below the horizontal does the ball approach the ground?

The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. If there is no air resistance, how long does it take the football to reach the top of its trajectory?

A projectile leaves the ground at 150 m/s and reaches a maximum height of 0.57 km. If there was no air resistance, at what angle above the horizontal did it leave the ground?

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 55° with the vertical. What is the length of the athlete's jump if air resistance is of no significance?

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s². The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively. At what angle was the projectile fired above the horizontal?

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s².

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s². What is the horizontal component of its velocity after 4.0 s?

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s². The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively. How long after firing does it take before the projectile hits the level ground?

In a room where g = 9.81 m/s², a hockey puck slides off the edge of a platform with an initial velocity of 28.0 m/s horizontally. The height of the platform above the ground is 2.00 m. What is the speed of the puck just before it hits the ground? The air resistance is negligibly small.