Question

You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly slips out of your hand and falls to the floor. Will it hit the elevator floor in more time, equal time, or less time than it would take if the elevator were standing still? Show work justifying your answer.

Answer 1

Answer:

It falls at the same speed in both cases.

Explanation:

If I were standing still the phone would be in free fall after slipping out of my hand.

I set a frame of reference with origin on the ground and the positive Y axis pointing up.

It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.

It would be subject to an gravitational acceleration of -32.2 ft/s^2.

Since acceleration is constant:

Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2

When it hits the floor at t1 it will be at Y(t1) = 0

0 = 5 + 0 * t1 - 16.1 * t1^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

If the elevator is standing still it would take 0.55 s to hit the ground.

Now, if the elevator is moving up at 10 ft/s.

The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t

Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.

And it will hit the floor of the elevator not at 0, but at

Ye = 10 * t2

So:

10 * t2 = 5 + 10 * t2 - 16.1 * t2^2

0 = 5 - 16.1 * t2^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

It falls at the same speed in both cases.