Question

At time t = 0 an elevator starts moving upward from the ground at a constant speed vo. At a later time t = T1 = 2.35s a marble is "let go" through the floor. In other words, imagine a marble resting on the floor of the elevator and a trap door of sorts opening up from under the marble. The marble then hits the ground at a still later time t = T2 = 11.2s.
1. What is the maximum height (above the ground) that the marble reaches in this process?

Answer 1

Answer:

$\frac{v_o^2}{19.62} + 2.35v_o$

Explanation:

The marble released through the trap door inside the elevator going upward at a speed vo would have an upward velocity of vo, but begins to experience a downward acceleration g = 9.81m/s^2. This is essentially same as throwing the marble up at a speed of vo. We can solve for the greatest height using the law of energy conservation.

When the marble is traveling up, its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv_o^2/2$

where m is the marble mass and h is the vertical distance traveled from the releasing point to the maximum height.

We can divide both sides by m

$gh = v_o^2/2$

$h = \frac{v_o^2}{2g} = \frac{v_0^2}{2*9.81} = \frac{v_0^2}{19.62}$

This is only partial of the height from the ground. To find the total height we need to add the height from the ground to the release point, which is the distance traveled by the elevator within time t = 2.35s at speed vo

$\frac{v_o^2}{19.62} + 2.35v_o$

Plug in the value for vo and we can calculate the maximum height