Question 1

A ball is projected upward at time t = 0 s,from a point on a flat roof 90 m above the ground.The ball rises and then falls with insignificant air resistance,missing the roof,and strikes the ground.The initial velocity of the ball is 80.5 m/s.Consider all quantities as positive in the upward direction.The vertical velocity of the ball when it is $89 \mathrm {~m}$ above the ground is closest to
A)-81 m/s.
B)-64 m/s.
C)-48 m/s.
D)-32 m/s.
E)-97 m/s.

Accepted Answer

The correct answer is A.Explanation:The vertical velocity of the ball when it is $89 \mathrm {~m}$ above the ground is closest to -81 m/s. This is because the ball is moving upward with a velocity of 80.5 m/s at time t = 0 s, and its velocity will decrease as it rises due to the force of gravity. The ball will reach its maximum height when its velocity is 0 m/s, and then it will fall back to the ground with a velocity that is equal in magnitude but opposite in direction to its initial velocity. Therefore, the vertical velocity of the ball when it is $89 \mathrm {~m}$ above the ground is closest to -81 m/s.

Question 2

A toy rocket is launched vertically from ground level 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 81 m and acquired an upward velocity of 40 m/s.The rocket continues to rise with insignificant air resistance in unpowered flight,reaches maximum height,and falls back to the ground.The upward acceleration of the rocket during the burn phase is closest to
A)9.9 m/ $\mathrm { s ^2}$
B)9.6 m/ $$\mathrm { s }^2$$
C)9.3 m/ $$\mathrm { s }^2$$
D)9.0 m/ $\mathrm { s }^2$
E)8.7 m/ $\mathrm { s }^2$

Accepted Answer

Let's start by using the kinematic equations for motion with constant acceleration. We'll take upward as positive.

During the burn phase:

Initial velocity, u = 0 
Final velocity, v = 40 m/s 
Displacement, s = 81 m 
Time, t = ?

We know that acceleration, a, is constant during the burn phase.

From the third kinematic equation:

v^2 = u^2 + 2as

40^2 = 2a(81)

a = 9.9 m/s^2

Therefore, the upward acceleration of the rocket during the burn phase is 9.9 m/s^2, which is closest to choice A).

Question 3

A package is dropped from a helicopter that is moving upward at 15 m/s If it takes $8.0 \mathrm {~s}$ before the package strikes the ground,how high above the ground was the package when it was released? Neglect air resistance.

Accepted Answer

The answer of A package is dropped from a helicopter...

Question 4

A bullet shot straight up returns to its starting point in 10 s.What is the initial speed of the bullet,assuming no air?
A)9.8 m/s
B)25 m/s
C)49 m/s
D)98 m/s

Accepted Answer

When a bullet is shot straight up, it will experience a constant downward acceleration due to gravity. The time taken for it to return back to the starting point is equal to the time taken for it to reach its maximum height and then fall back down. Therefore, the total time of flight is 10 s, which means it took 5 s to reach the maximum height.

Using the kinematic equation, 
Δy = v0t + (1/2)at²

where Δy is the displacement (which is zero as the bullet returns to its initial point), v0 is the initial velocity that we need to find, t is the time of flight (5 s to reach maximum height), and a is the acceleration due to gravity (-9.8 m/s²),

0 = v0(5) + (1/2)(-9.8)(5)²
v0 = 49 m/s

Hence, the initial speed of the bullet is 49 m/s. Therefore, the correct answer is choice C.

Question 5

A ball is thrown straight up with a speed of 30 m/s,and you can neglect the presence of air.
(a)How long does it take the ball to reach the maximum height?
(b)What is the maximum height reached by the ball?
(c)What is its speed after 4.2 s?

Accepted Answer

The answer of A ball is thrown straight up with...

Question 6

A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall.The rock misses the building on its way down,and is observed to strike the ground 4.00 seconds after being thrown.Take the acceleration due to gravity to have magnitude 9.80 m/s² and 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 7

A ball is thrown downward from the top of a building with an initial speed of 25 m/s.It strikes the ground after 2.0 s.How high is the building,assuming no air?
A)20 m
B)30 m
C)50 m
D)70 m

Accepted Answer

We can use the kinematic equation for vertical motion with constant acceleration (due to gravity):

h = vi*t + (1/2)*g*t^2

Where:
h = Height of the building
vi = Initial velocity (upward) = -25 m/s (because the ball is thrown downward)
t = Time taken to hit the ground = 2.0 s
g = Acceleration due to gravity = -9.8 m/s^2 (downward)

Substituting the given values, we get:

h = -25*2 + (1/2)*(-9.8)*(2^2) = 70 m

Therefore, the height of the building is 70 m. The correct answer is D.

Question 8

To determine the height of a bridge above the water,a person drops a stone and measures the time it takes for it to hit the water.If the height of the bridge is 41 m,how long will it take for the stone to hit the water? Neglect air resistance.
A)2.3 s
B)2.6 s
C)2.9 s
D)3.2 s
E)3.6 s

Accepted Answer

The time it takes for an object to fall a certain distance under the influence of gravity alone (neglecting air resistance) can be calculated using the formula $t = \sqrt{\frac{2h}{g}}$, where $t$ is the time in seconds, $h$ is the height in meters, and $g$ is the acceleration due to gravity, approximately $9.8 \, \text{m/s}^2$. Plugging in the given height of 41 meters, we get $t = \sqrt{\frac{2 \times 41}{9.8}} \approx 2.9$ seconds.

Question 9

A ball is thrown straight upward from ground level with a speed of 18 m/s.How much time passes before the ball strikes the ground if we disregard air resistance?
A)3.7 s
B)1.8 s
C)1.1 s
D)0.6 s

Accepted Answer

The time for the ball to reach its highest point can be calculated using the formula $t = \frac{v}{g}$, where $v = 18 \, \text{m/s}$ is the initial velocity and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity. This gives $t = \frac{18}{9.8} \approx 1.84 \, \text{s}$. Since the time to go up is the same as the time to come down, the total time is $2 \times 1.84 \approx 3.7 \, \text{s}$.

Question 10

An object is dropped from a bridge.A second object is thrown downwards 1.0 s later.They both reach the water 20 m below at the same instant.What was the initial speed of the second object? Neglect air resistance.
A)4.9 m/s
B)15 m/s
C)9.9 m/s
D)20 m/s
E)21 m/s

Accepted Answer

The answer of An object is dropped from a bridge.A...

Question 11

To determine the height of a bridge above the water,a person drops a stone and measures the time it takes for it to hit the water.If the time is 2.3 s,what is the height of the bridge? Neglect air resistance.
A)10 m
B)14 m
C)26 m
D)32 m
E)52 m

Accepted Answer

The answer of To determine the height of a bridge...

Question 12

A toy rocket is launched vertically from ground level at time t = 0 s.The rocket engine provides constant upward acceleration during the burn phase.At the instant of engine burnout,the rocket has risen to 49.0 m and acquired an upward velocity of 60.0 m/s.The rocket continues to rise with insignificant air resistance in unpowered flight,reaches maximum height,and falls back to the ground.The maximum height reached by the rocket is closest to
A)233 m.
B)221 m.
C)209 m.
D)244 m.
E)256 m.

Accepted Answer

To find the maximum height reached by the rocket, we use the equation of motion under constant acceleration, $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m/s at the maximum height), $u$ is the initial velocity (60.0 m/s), $a$ is the acceleration due to gravity (-9.8 m/s$^2$, negative because it's acting opposite to the direction of motion), and $s$ is the displacement (the additional height gained after the engine burnout). Rearranging the equation for $s$, we get $s = \frac{v^2 - u^2}{2a}$. Substituting the given values, $s = \frac{0 - (60.0)^2}{2(-9.8)} = \frac{-3600}{-19.6} = 183.67 m$. This is the height gained after the engine burnout. The total height reached is the sum of the height at burnout (49.0 m) and the additional height gained (183.67 m), which equals approximately 232.67 m, which is closest to 233 m.

Question 13

A ball is thrown straight up with a speed of 36 m/s.How long does it take to return to its starting point,assuming no air?
A)3.7 s
B)7.3 s
C)11 s
D)15 s

Accepted Answer

The total time for the ball to return to its starting point can be found using the formula $t = \frac{2v}{g}$, where $v$ is the initial velocity (36 m/s) and $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$). Plugging in the values, $t = \frac{2 \times 36}{9.8} \approx 7.3$ seconds.

Question 14

A rock is projected upward from the surface of the Moon,at time t = 0 s,with an upward velocity of 30.0 m/s.The acceleration due to gravity at the surface of the Moon is 1.62 m/s²,and the Moon has no atmosphere.The height of the rock when it is descending with a speed of 20.0 m/s is closest to A)115 m. B)125 m. C)135 m. D)145 m. E)154 m.

Accepted Answer

We can use the kinematic equation:

v_f^2 = v_i^2 + 2ad

where v_f is the final velocity (20.0 m/s), v_i is the initial velocity (30.0 m/s), a is the acceleration due to gravity on the Moon (-1.62 m/s^2), and d is the distance traveled.

Solving for d, we get:

d = (v_f^2 - v_i^2)/(2a)
  = (20.0^2 - 30.0^2)/(2*(-1.62))
  = 154.32 m

Therefore, the height of the rock when it is descending with a speed of 20.0 m/s is closest to 154 m. The best choice is E.

Question 15

A ball is thrown upward at a velocity of 19.6 m/s.What is its velocity after 3.0 s,assuming no air?
A)9.8 m/s upward
B)9.8 m/s downward
C)0 m/s
D)19.6 m/s downward

Accepted Answer

The acceleration due to gravity is 9.8 m/s^2 downward. After 3 seconds, the velocity change due to gravity is 9.8 m/s^2 * 3 s = 29.4 m/s. Since the initial velocity is 19.6 m/s upward, the final velocity is 9.8 m/s downward (19.6 m/s upward - 29.4 m/s = -9.8 m/s, indicating a downward direction).

Question 16

A foul ball is hit straight up into the air with a speed of 30 m/s,and air effects are negligible.
(a)Calculate the time required for the ball to rise to its maximum height.
(b)Calculate the maximum height reached by the ball above the point where it hit the bat.
(c)Determine the times at which the ball passes a point 25 m above the point where it was hit by the bat.
(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 ball is projected upward at time t = 0 s,from a point on a flat roof 10 m above the ground.The ball rises and then falls,missing the roof,and strikes the ground.The initial velocity of the ball is $58.5 \mathrm {~m} / \mathrm { s }$ Consider all quantities as positive in the upward direction.Neglect the effects of air.At time $t = 5.97 \mathrm {~s} ,$ the vertical velocity of the ball is closest to
A)0 m/s.
B)+175 m/s.
C)+12 m/s.
D)-175 m/s.
E)-12 m/s.

Accepted Answer

At the highest point of the trajectory, the vertical velocity of the ball is 0 m/s.

Question 18

A test rocket at ground level is fired straight up from rest with a net upward acceleration of 20 m/s².After 4.0 s,the motor turns off but the rocket continues to coast upward with insignificant air resistance.What maximum elevation does the rocket reach? A)160 m B)330 m C)320 m D)410 m E)490 m

Accepted Answer

The answer of A test rocket at ground level is...

Question 19

At the same moment,one rock is dropped and one is thrown downward with an initial velocity of $29 \mathrm {~m} / \mathrm { s }$ from the top of a building that is 300 m tall.How much earlier does the thrown rock strike the ground? Neglect air resistance.

Accepted Answer

The answer of At the same moment,one rock is dropped...

Question 20

A toy rocket is launched vertically from ground level 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 64 m and acquired an upward velocity of 60 m/s.The rocket continues to rise with insignificant air resistance in unpowered flight,reaches maximum height,and falls back to the ground.The time interval during which the rocket engine provided the upward acceleration,is closest to
A)2.1 s.
B)2.3 s.
C)1.9 s.
D)1.7 s.
E)1.5 s.

Accepted Answer

The correct answer is A because the rocket's initial velocity is 0 m/s, and its final velocity at burnout is 60 m/s. The acceleration during the burn phase is therefore (60 m/s - 0 m/s) / 2.1 s = 28.57 m/s^2. The time interval during which the rocket engine provided the upward acceleration is therefore 2.1 s.

Quiz 2: Kinematics: Motion in One Dimension

A package is dropped from a helicopter that is moving upward at 15 m/s If it takes 8.0 s8.0 \mathrm {~s}8.0 s before the package strikes the ground,how high above the ground was the package when it was released? Neglect air resistance.

A bullet shot straight up returns to its starting point in 10 s.What is the initial speed of the bullet,assuming no air?

A ball is thrown straight up with a speed of 30 m/s,and you can neglect the presence of air. (a)How long does it take the ball to reach the maximum height? (b)What is the maximum height reached by the ball? (c)What is its speed after 4.2 s?

A ball is thrown downward from the top of a building with an initial speed of 25 m/s.It strikes the ground after 2.0 s.How high is the building,assuming no air?

To determine the height of a bridge above the water,a person drops a stone and measures the time it takes for it to hit the water.If the height of the bridge is 41 m,how long will it take for the stone to hit the water? Neglect air resistance.

A ball is thrown straight upward from ground level with a speed of 18 m/s.How much time passes before the ball strikes the ground if we disregard air resistance?

An object is dropped from a bridge.A second object is thrown downwards 1.0 s later.They both reach the water 20 m below at the same instant.What was the initial speed of the second object? Neglect air resistance.

To determine the height of a bridge above the water,a person drops a stone and measures the time it takes for it to hit the water.If the time is 2.3 s,what is the height of the bridge? Neglect air resistance.

A ball is thrown straight up with a speed of 36 m/s.How long does it take to return to its starting point,assuming no air?

A rock is projected upward from the surface of the Moon,at time t = 0 s,with an upward velocity of 30.0 m/s.The acceleration due to gravity at the surface of the Moon is 1.62 m/s2,and the Moon has no atmosphere.The height of the rock when it is descending with a speed of 20.0 m/s is closest to

A ball is thrown upward at a velocity of 19.6 m/s.What is its velocity after 3.0 s,assuming no air?

A test rocket at ground level is fired straight up from rest with a net upward acceleration of 20 m/s2.After 4.0 s,the motor turns off but the rocket continues to coast upward with insignificant air resistance.What maximum elevation does the rocket reach?

At the same moment,one rock is dropped and one is thrown downward with an initial velocity of 29 m/s29 \mathrm {~m} / \mathrm { s }29 m/s from the top of a building that is 300 m tall.How much earlier does the thrown rock strike the ground? Neglect air resistance.

A package is dropped from a helicopter that is moving upward at 15 m/s If it takes $8.0 \mathrm {~s}$ before the package strikes the ground,how high above the ground was the package when it was released? Neglect air resistance.

A rock is projected upward from the surface of the Moon,at time t = 0 s,with an upward velocity of 30.0 m/s.The acceleration due to gravity at the surface of the Moon is 1.62 m/s²,and the Moon has no atmosphere.The height of the rock when it is descending with a speed of 20.0 m/s is closest to

A test rocket at ground level is fired straight up from rest with a net upward acceleration of 20 m/s².After 4.0 s,the motor turns off but the rocket continues to coast upward with insignificant air resistance.What maximum elevation does the rocket reach?

At the same moment,one rock is dropped and one is thrown downward with an initial velocity of $29 \mathrm {~m} / \mathrm { s }$ from the top of a building that is 300 m tall.How much earlier does the thrown rock strike the ground? Neglect air resistance.