Question 1

A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal. How far from the foot of the vertical cliff does the projectile hit the water?

Accepted Answer

The correct answer is A because the projectile's horizontal displacement is given by:$$x = v_0 \cos(	heta) t$$where $$v_0$$ is the initial velocity, $$	heta$$ is the angle below the horizontal, and t is the time of flight. The time of flight is given by:$$t = \frac{2v_0 \sin(	heta)}{g}$$where g is the acceleration due to gravity. Substituting these expressions into the equation for x, we get:$$x = v_0 \cos(	heta) \frac{2v_0 \sin(	heta)}{g}$$$$x = \frac{v_0^2 \sin(2	heta)}{g}$$Plugging in the given values, we get:$$x = \frac{(100 \ m/s)^2 \sin(2(35.0°))}{9.81 \ m/s^2}$$$$x = 79.1 \ m$$

Question 2

A ball is thrown at an angle θ above the horizontal. In the presence of the acceleration due to gravity, how far below the straight-line path following its original direction of motion will the ball be at the end of 1.0 second?

Accepted Answer

The vertical displacement due to gravity after 1 second is given by the formula $ \frac{1}{2}gt^2 $, where $ g = 9.8 \, \text{m/s}^2 $. Substituting $ t = 1 \, \text{s} $, the displacement is $ \frac{1}{2} \times 9.8 \times 1^2 = 4.9 \, \text{m} $.

Question 3

A bullet is fired from ground level with a speed of 150 m/s at an angle 30.0° above the horizontal at a location where g = 10.0 m/s². What is the vertical component of its velocity after 4 seconds?

Accepted Answer

The answer of A bullet is fired from ground level...

Question 4

A bullet is fired from ground level with a speed of 150 m/s at an angle 30.0° above the horizontal at a location where g = 10.0 m/s². What is the horizontal component of its velocity after 4 seconds?

Accepted Answer

First, we need to find the initial horizontal and vertical components of the bullet's velocity.

The initial horizontal velocity is given by:

v_x = v0cos(theta) = 150*cos(30) = 129.9 m/s

The initial vertical velocity is given by:

v_y = v0sin(theta) = 150*sin(30) = 75.0 m/s

Next, we need to find how far the bullet has traveled horizontally after 4 seconds. We can do this by using the equation:

x = v_x*t = 129.9*4 = 519.6 m

Now, we want to find the horizontal component of the bullet's velocity after 4 seconds, which is simply the initial horizontal velocity:

v_x = 129.9 m/s

Therefore, the best choice is C) 130 m/s.

Question 5

A ball is thrown horizontally from the top of a building. What will be its vertical displacement from the top of building 2.0 seconds after it leaves your hands?

Accepted Answer

The vertical displacement of the ball can be calculated using the formula d = (1/2)gt^2, where d is the displacement, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Plugging in the values, we get: 
d = (1/2)(9.8 m/s^2)(2.0 s)^2
d = 19.6 m
Therefore, the vertical displacement of the ball 2.0 seconds after it leaves your hands is 19.6 meters.

Question 6

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. How far from where the child is standing does the ball hit the ground?

Accepted Answer

The answer of A child throws a ball with an...

Question 7

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. At what angle below the horizontal does the ball approach the ground?

Accepted Answer

The answer of A child throws a ball with an...

Question 8

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. How long is the ball in flight before it hits the ground?

Accepted Answer

We can use the kinematic equations to solve for the time the ball is in the air. Since we know the initial velocity of the ball, we can use the y-component of the velocity to find how long it takes for the ball to reach its maximum height.

vy = voy + at 
0 = 8.00 sin(40.0) - 9.81t 
t = 1.22 s

Since the time to reach the maximum height is the same as the time to fall back to the ground, the total time the ball is in the air is twice as long:

t_total = 2t 
t_total = 2 x 1.22 
t_total = 2.44 s

Therefore, the answer is A) 1.22 s.

Question 9

An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplane to land on a target on the ground. At what horizontal distance away from the target should the package be released so that it lands on the target?

Accepted Answer

The correct answer is A because the horizontal distance away from the target should be 2.55 km so that the package lands on the target.

Question 10

An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplane to land on a target on the ground. In what direction will the package be traveling when it hits the ground?

Accepted Answer

The answer of An airplane is flying at a speed...

Question 11

A person throws a ball horizontally from the top of a building that is 40.0 m high. The initial velocity of the ball is 100 m/s. What is the horizontal distance that the ball travels before hitting the ground?

Accepted Answer

The horizontal distance traveled by the ball is given by:```d = v*t```where:* d is the distance traveled (in meters)* v is the initial velocity (in meters per second)* t is the time of flight (in seconds)The time of flight can be found using the following equation:```t = sqrt(2*h/g)```where:* h is the height of the building (in meters)* g is the acceleration due to gravity (in meters per second squared)Substituting the given values into these equations, we get:```t = sqrt(2*40.0 m/9.8 m/s^2) = 4.08 sd = 100 m/s * 4.08 s = 286 m```Therefore, the horizontal distance that the ball travels before hitting the ground is 286 m.

Question 12

A ball is thrown horizontally with an initial velocity of 20.0 m/s from the edge of a building of a certain height. The ball lands at a horizontal distance of 82.0 m from the base of the building. What is the height of the building?

Accepted Answer

The answer of A ball is thrown horizontally with an...

Question 13

A boy kicks a football with a certain initial velocity at an angle 20° above the horizontal. In 2.0 seconds, the ball reaches at its highest point in its trajectory. What is the initial velocity of the ball?

Accepted Answer

Using kinematic equations, we know that the time it takes for an object to reach its highest point in its trajectory is given by:
t = (Vy_f - Vy_i) / g
where t = 2.0 seconds, Vy_f = 0 m/s (at the highest point, the vertical velocity is zero), Vy_i = V0 sin(20°) (initial vertical velocity), and g = 9.8 m/s^2 (acceleration due to gravity). 
Solving for V0, we get:
V0 = Vy_i / sin(20°) = (g*t)/sin(20°) = (9.8 m/s^2 * 2.0 s) / sin(20°) ≈ 57 m/s. 
Therefore, the initial velocity of the ball was 57 m/s.

Question 14

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

Accepted Answer

The correct answer is A because the ball is at its maximum height after 2.0 s. The vertical component of the initial velocity is 25 m/s * sin(30°) = 12.5 m/s. The acceleration due to gravity is -9.8 m/s^2. The time it takes for the ball to reach its maximum height is 12.5 m/s / 9.8 m/s^2 = 1.28 s. The maximum height is 12.5 m/s * 1.28 s - 0.5 * 9.8 m/s^2 * (1.28 s)^2 = 5.4 m.

Question 15

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and the fence is 2.00 m high. The ball just clears the fence while still traveling upwards. How far is the child from the fence?

Accepted Answer

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

Question 16

A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0° above the horizontal. How far from the foot of the vertical cliff does the projectile hit the water?

Accepted Answer

The answer of A projectile is shot from the edge...

Question 17

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?

Accepted Answer

The magnitude of the ball's velocity just before it hits the ground can be found using energy conservation or kinematic equations. The initial kinetic energy and potential energy are converted into kinetic energy just before impact. The velocity can be calculated using the formula: v = sqrt(vx^2 + vy^2), where vx = v0 * cos(θ) and vy = sqrt(v0^2 * sin^2(θ) + 2gh). Solving gives approximately 9.14 m/s.

Question 18

A ball is thrown with an initial speed of 60 m/s at an angle of 30° above the horizontal. What is the ball's horizontal displacement at the end of 4 seconds? Use g = 10 m/s².

Accepted Answer

The answer of A ball is thrown with an initial...

Question 19

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 vertical component of its velocity at the highest point of its trajectory?

Accepted Answer

At the highest point of its trajectory, the vertical component of the velocity is 0 m/s because the ball momentarily stops moving upward before descending.

Question 20

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?

Accepted Answer

The answer of A person throws a ball horizontally from...

Quiz 4: Two-Dimensional Kinematics