Question

A ball is released from the bottom of an elevator which is traveling upward with a velocity of 1.828 m/s. If the ball strikes the bottom of the elevator shaft in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released. Also, find the velocity of the ball when it strikes the bottom of the shaft.​

Answer 1

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Answer 2

Answer:

The velocity of the ball when it strikes the bottom of the shaft is 27.916 m/s downward.

Explanation:

Let's start by finding the initial velocity of the ball when it is released from the elevator. Since the elevator is traveling upward with a velocity of 1.828 m/s, the initial velocity of the ball relative to the ground is also 1.828 m/s upward.

Now, we can use the kinematic equation for vertical motion:

h = vit + 1/2gt^2

where:

h = height of the elevator from the bottom of the shaft

vi = initial velocity of the ball (1.828 m/s upward)

g = acceleration due to gravity (-9.8 m/s^2)

t = time taken by the ball to strike the bottom of the shaft (3 s)

Substituting the values, we get:

h = (1.828 m/s)(3 s) + 1/2(-9.8 m/s^2)(3 s)^2

h = 5.484 m - 44.1 m

h = -38.616 m

The negative sign indicates that the elevator is actually below the bottom of the shaft. Therefore, the height of the elevator from the bottom of the shaft at the instant the ball is released is 38.616 m below the bottom of the shaft.

To find the velocity of the ball when it strikes the bottom of the shaft, we can use the same kinematic equation:

v = vi + gt

where:

v = final velocity of the ball

vi = initial velocity of the ball (1.828 m/s upward)

g = acceleration due to gravity (-9.8 m/s^2)

t = time taken by the ball to strike the bottom of the shaft (3 s)

Substituting the values, we get:

v = 1.828 m/s - 9.8 m/s^2 (3 s)

v = -27.916 m/s

The negative sign indicates that the velocity of the ball is downward. Therefore, the velocity of the ball when it strikes the bottom of the shaft is 27.916 m/s downward.

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