Question

Crall decides that with the kinematic equations of motion and the value of g (9.81 m/s2) she should be able to make quantitative predictions of the ball's motion. She decides to release the ball at a height of 40 m above the ground and that she wants the ball to hit the ground at time 4 s later. She then calculates the speed the elevator needs to be traveling in order for the ball to reach the ground at this time. Which is the correct result of Crall's calculation? −9.6 m/s 29.6 m/s

Answer 1

V_avg = -9.6 m/s

Step-by-step explanation:

Given:

- Initial height y(0) = 40 m

- Initial velocity v(0) = 0

- Time taken for the ball to reach ground t = 4 s

Find:

- The speed of the elevator for it to reach the ball in time:

Solution:

- For the elevator to reach the ground within 4 s seconds till the ball hits the ground we can calculate and average constant speed of the elevator, assuming the elevator does not accelerates:

V_avg = ( y(f) - y(0) ) / dt

V_avg = (0 - 40) / 4

V_avg = -40/4

V_avg = -10 m/s

- If assuming the elevator moves with a constant acceleration down, then then Crall's calculation of -9.6 m/s is correct.

Answer 2

u = 29.6 m/s

Step-by-step explanation:

Crall decides that with the kinematic equations of motion. She released the ball from the height of 40 meters and she wants the ball to hit the ground at time 4 s later.

Let v is the speed of the elevator needs to be traveling in order for the ball to reach the ground at this time. It can be solved using the second equation of motion as :

`h=ut+(1)/(2)at^2`

Here, a = -g

`h=ut-(1)/(2)gt^2`

`u=(h+(1)/(2)gt^2)/(t)`

`u=(40+(1)/(2)* 9.8* 4^2)/(4)`

u = 29.6 m/s

So, the speed the elevator needs to be traveling in order for the ball to reach the ground at this time is 29.6 m/s. Hence, this is the required solution.