Question 1

A cannonball at ground level is aimed 30.0° above the horizontal and is fired with an initial speed of 125 m/s. How far away from the cannon will the cannonball hit the ground?
A)125 m
B)138 m
C)695 m
D)1040 m
E)1380 m

Accepted Answer

The horizontal range of a projectile is given by the formula R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity (approximately 9.81 m/s²). Substituting the given values: R = (125^2 * sin(60°)) / 9.81 ≈ 1380 m.

Question 2

A spring-loaded gun is aimed horizontally and is used to launch identical balls with different initial speeds. The gun is at a fixed position above the floor. The balls are fired one at a time.  If the speed of the second ball fired is twice the speed of the first ball fired, how is the horizontal range (denoted R in the figure) affected?
A)The range for both balls will be the same.
B)The range of the second ball will be half as much as that of the first ball.
C)The range of the second ball will be twice as large as that of the first ball.
D)The range of the second ball is about 1.4 times larger than that of the first ball.
E)The range of the second ball will be smaller than that of the first ball by a factor of 1.4.

Accepted Answer

The answer of A spring-loaded gun is aimed horizontally and...

Question 3

A bullet is aimed at a target on the wall a distance L away from the firing position. Because of gravity, the bullet strikes the wall a distance $\Delta$y below the mark as suggested in the figure. Note: The drawing is not to scale.  If the distance L was half as large and the bullet had the same initial velocity, how would $\Delta$y be affected?
A)$\Delta$y will double.
B)$\Delta$y will be half as large.
C)$\Delta$y will be four times larger.
D)$\Delta$y will be one fourth as large.
E)$\Delta$y is not possible to determine unless numerical values are given for the distances.

Accepted Answer

The answer of A bullet is aimed at a target...

Question 4

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale.

-Determine the initial speed of the shell.
A)4.0 m/s
B)9.8 m/s
C)82 m/s
D)170 m/s
E)330 m/s

Accepted Answer

The answer of A shell is fired with a horizontal...

Question 5

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -What is the acceleration of the projectile when it reaches its maximum height? A)zero m/s² B)9.8 m/s², downward C)4.9 m/s², downward D)less than 9.8 m/s²and non-zero. E)Its magnitude is 9.8 m/s²;and its direction is changing.

Accepted Answer

The answer of A projectile fired from a gun has...

Question 6

A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the basket 0.96 s after it is launched. What are the distances x and y? Note: The drawing is not to scale.  x y
A)5.4 m 0.73 m
B)7.7 m 0.91 m
C)5.4 m 0.91 m
D)5.7 m 0.73 m
E)5.4 m 5.4 m

Accepted Answer

Using the kinematic equations:
Vertical motion (y-axis):
y = vt + 1/2at^2
where y = 0.91 m (height of the basket), v = 8.0 m/s (initial velocity), and t = 0.96 s (time of flight).
0.91 m = (8.0 m/s)(0.96 s) + 1/2(9.81 m/s^2)(0.96 s)^2
0.91 m = 7.68 m + 4.72 m
0.91 m = 12.4 m
Therefore, the vertical distance traveled is 0.91 m.

Horizontal motion (x-axis):
x = vt
where x is the horizontal distance traveled, v is the initial velocity, and t is the time of flight (0.96 s).
x = (8.0 m/s)(0.96 s)
x = 7.68 m
Therefore, the horizontal distance traveled is 7.68 m.
The answer is (C) 5.4 m 0.91 m.

Question 7

A projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s.
-What is the magnitude of the horizontal component of the projectile's displacement at the end of 2 s?
A)30 m
B)40 m
C)10 m
D)20 m
E)50 m

Accepted Answer

The horizontal component of the initial velocity is given by $ v_{x} = v \cdot \cos(\theta) $. Here, $ v = 35.0 \, \text{m/s} $ and $ \theta = 55.0^\circ $. So, $ v_{x} = 35.0 \cdot \cos(55.0^\circ) \approx 20.1 \, \text{m/s} $. The horizontal displacement after 2 seconds is $ v_{x} \cdot t = 20.1 \, \text{m/s} \times 2 \, \text{s} \approx 40.2 \, \text{m} $, which is closest to 40 m.

Question 8

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
-Approximately how long does it take the projectile to reach the highest point in its trajectory?
A)1 s
B)2 s
C)4 s
D)8 s
E)16 s

Accepted Answer

The answer of A projectile fired from a gun has...

Question 9

A puck slides across a smooth, level tabletop at height H at a constant speed v_o. It slides off the edge of the table and hits the floor a distance x away as shown in the figure. What is the relationship between the distances x and H? A) B) C) D) E)

Accepted Answer

The horizontal motion of the puck is not affected by its height above the table, so it will continue to move at a constant speed v_o once it leaves the table. The only force acting on it after leaving the table is gravity, which will cause the puck to accelerate downward at a rate of g. Using the equation of motion for vertical motion, we can find the time it takes for the puck to hit the ground: H = (1/2)gt^2 t = sqrt(2H/g) Once we know the time, we can use the horizontal speed and the time to find the distance x that the puck travels before hitting the ground: x = v_o * t Substituting the expression for t: x = v_o * sqrt(2H/g) We can see that x is proportional to sqrt(H), so the correct relationship between x and H is: x ∝ sqrt(H) or x = k * sqrt(H) where k is some constant. This means that x will increase as the square root of H increases. Therefore, the answer is choice A.

Question 10

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
-At what angle is the projectile fired (measured with respect to the horizontal)?
A)37°
B)40°
C)45°
D)53°
E)60°

Accepted Answer

The angle θ can be found using the tangent function: tan(θ) = vertical velocity / horizontal velocity = 40/30. Solving for θ gives θ = arctan(4/3), which is approximately 53°.

Question 11

A tennis ball is thrown upward at an angle from point A. It follows a parabolic trajectory and hits the ground at point D. At the instant shown, the ball is at point B. Point C represents the highest 
-While in flight, how do the x and y components of the velocity vector of the ball compare at the points B and C?
A)The velocity components are non-zero at B and zero at C.
B)The x components are the same; the y component at C is zero m/s.
C)The x components are the same; the y component has a larger magnitude at C than at B.
D)The x component is larger at C than at B; the y component at B points up while at C, it points downward.
E)The x component is larger at B than at C; the y component at B points down while at C, it points upward.

Accepted Answer

In projectile motion, the horizontal (x) component of velocity remains constant, while the vertical (y) component changes due to gravity. At the peak (point C), the y component is zero.

Question 12

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
-Determine the initial speed of the projectile.
A)40 m/s
B)50 m/s
C)60 m/s
D)70 m/s
E)80 m/s

Accepted Answer

The initial speed of the projectile can be found using the Pythagorean theorem:

initial speed = √(horizontal velocity² + vertical velocity²)
initial speed = √(30² + 40²)
initial speed = √(900 + 1600)
initial speed = √2500
initial speed = 50 m/s

Therefore, the correct answer is B.

Question 13

A projectile is fired horizontally with an initial speed of 57 m/s. What are the horizontal and vertical components of its displacement 3.0 s after it is fired? horizontal vertical
A)44 m -29 m
B)170 m zero m
C)170 m -44 m
D)210 m -44 m
E)210 m zero m

Accepted Answer

The horizontal component of displacement is given by:

distance = speed x time = 57 m/s x 3.0 s = 170 m (since there is no acceleration in the horizontal direction)

The vertical component of displacement is given by:

distance = initial velocity x time + 0.5 x acceleration x time^2

where the initial velocity is zero (since the projectile is fired horizontally) and the acceleration is -9.8 m/s^2 (since the projectile is subject to gravity).

distance = 0 x 3.0 s + 0.5 x (-9.8 m/s^2) x (3.0 s)^2 = -44.1 m (the negative sign indicates the displacement is in the downward direction)

Therefore, the answer is C: 170 m horizontal displacement and -44 m vertical displacement.

Question 14

A projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s.
-How long does it take the projectile to reach the highest point in its trajectory?
A)1.5 s
B)2.4 s
C)4.0 s
D)6.2 s
E)9.8 s

Accepted Answer

The time it takes for the projectile to reach the highest point in its trajectory can be calculated using the formula: 
time = (vertical initial velocity) / (gravitational acceleration) 
At the highest point, the vertical initial velocity is zero, so we can use: 
time = (vertical final velocity) / (gravitational acceleration) 
The vertical final velocity at the highest point can be calculated using: 
vertical final velocity = (initial velocity) * sin(angle) 
where angle is the angle of the initial velocity vector with the horizontal. 
Using the given values, we get: 
vertical final velocity = (35.0 m/s) * sin(55.0°) ≈ 28.4 m/s 
time = (28.4 m/s) / (9.81 m/s^2) ≈ 2.89 s 
Therefore, the answer is B) 2.4 s, which is the closest choice to our calculation.

Question 15

An arrow is shot horizontally from a height of 4.9 m above the ground. The initial speed of the arrow is 45 m/s. Neglecting friction, how long will it take the arrow to hit the ground?
A)9.2 s
B)6.0 s
C)1.0 s
D)1.4 s
E)4.6 s

Accepted Answer

The answer of An arrow is shot horizontally from a...

Question 16

What is the magnitude of the projectile's velocity just before it strikes the ground?
A)zero m/s
B)9.8 m/s
C)30 m/s
D)40 m/s
E)50 m/s

Accepted Answer

The magnitude of the projectile's velocity just before it strikes the ground is typically the same as its initial velocity if air resistance is negligible and it was launched from the ground.

Question 17

A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later. Determine the distance the ball was thrown.
A)13 m
B)18 m
C)36 m
D)52 m
E)72 m

Accepted Answer

First, we need to find the horizontal and vertical components of the velocity:

vx = vcosθ = 25cos35° ≈ 20.5 m/s
vy = vsinθ = 25sin35° ≈ 14.4 m/s

Then, we can use the vertical motion equation to find the time it takes for the ball to reach its highest point:

vy = voy - gt
0 = 14.4 - 9.8t
t = 1.47 s

Since the total time in the air is 2.55 s, the time the ball spends falling from its highest point to the receiver's hands is:

t_fall = 2.55 s - 1.47 s = 1.08 s

During this time, the ball will fall a distance equal to:

Δy = voyt_fall - (1/2)gt_fall^2
Δy = 14.4(1.08) - (1/2)(9.8)(1.08)^2
Δy ≈ 6.4 m

Therefore, the receiver must have been 6.4 m below the quarterback when he caught the ball. To find the total distance the ball traveled, we can use the horizontal component of the velocity:

d = vx*t_total
d = 20.5*2.55
d ≈ 52 m

So the answer is D, 52 m.

Question 18

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale.

-What is the magnitude of the velocity of the shell as it hits the ground?
A)4.0 m/s
B)9.8 m/s
C)82 m/s
D)170 m/s
E)330 m/s

Accepted Answer

The answer of A shell is fired with a horizontal...

Question 19

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
-What is the speed of the projectile when it is at the highest point in its trajectory?
A)0 m/s
B)20 m/s
C)30 m/s
D)40 m/s
E)50 m/s

Accepted Answer

At the highest point of the trajectory, the vertical component of velocity is zero. This means that the projectile has lost its vertical speed and is moving only in the horizontal direction at a speed of 30 m/s. Since the horizontal component of velocity remains constant, the speed of the projectile at the highest point is equal to the horizontal component of velocity, which is 30 m/s. Therefore, the correct choice is C) 30 m/s.

Question 20

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. -What is the magnitude of the acceleration of the shell just before it strikes the ground? A)4.0 m/s² B)9.8 m/s² C)82 m/s² D)170 m/s² E)330 m/s²

Accepted Answer

The horizontal velocity is constant, so we can use the horizontal distance and time of flight to find the horizontal velocity.

d = vt
1330 m = v(t)
t = 1330 m/v

The vertical distance can be found using the equation:
y = vi(t) + (1/2)at^2
where vi = 0 and y = -80 m (taking upward as positive).

-80 m = (1/2)(a)(t^2)
a = - (2*80 m) / (t^2)

Substituting t from the first equation:
a = - (2*80 m) / ((1330 m/v)^2)

The negative sign indicates that the acceleration is downward (towards the ground).

Now, using the equation for range:
d = ((v^2)/g) * sin(2θ)

Since the shell was fired horizontally, θ = 0, so:
v = sqrt((d*g)/sin(2θ)) = sqrt((1330 m/s^2) / sin(0)) = 115.4 m/s

Therefore:
a = - (2*80 m) / ((1330 m/115.4 m/s)^2) = -9.8 m/s^2

The magnitude of this acceleration is 9.8 m/s^2, which corresponds to answer choice B.

Quiz 3: Kinematics in Two Dimensions

A cannonball at ground level is aimed 30.0° above the horizontal and is fired with an initial speed of 125 m/s. How far away from the cannon will the cannonball hit the ground?

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. -Determine the initial speed of the shell.

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -What is the acceleration of the projectile when it reaches its maximum height?

A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the basket 0.96 s after it is launched. What are the distances x and y? Note: The drawing is not to scale. x y

A projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s. -What is the magnitude of the horizontal component of the projectile's displacement at the end of 2 s?

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -Approximately how long does it take the projectile to reach the highest point in its trajectory?

A puck slides across a smooth, level tabletop at height H at a constant speed v_o. It slides off the edge of the table and hits the floor a distance x away as shown in the figure. What is the relationship between the distances x and H?

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -At what angle is the projectile fired (measured with respect to the horizontal) ?

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -Determine the initial speed of the projectile.

A projectile is fired horizontally with an initial speed of 57 m/s. What are the horizontal and vertical components of its displacement 3.0 s after it is fired? horizontal vertical

A projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s. -How long does it take the projectile to reach the highest point in its trajectory?

An arrow is shot horizontally from a height of 4.9 m above the ground. The initial speed of the arrow is 45 m/s. Neglecting friction, how long will it take the arrow to hit the ground?

What is the magnitude of the projectile's velocity just before it strikes the ground?

A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later. Determine the distance the ball was thrown.

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. -What is the magnitude of the velocity of the shell as it hits the ground?

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. -What is the speed of the projectile when it is at the highest point in its trajectory?

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. -What is the magnitude of the acceleration of the shell just before it strikes the ground?