A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the ba
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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
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
It falls at the same speed in both cases.