Question

An elevator ascends from the ground with uniform speed. A time Tį later, a boy drops a marble through a hole in the floor. A time T2 after that (i.e. Ti +T2 after start) the marble hits the ground. Find an expression for the height of the elevator at time Ti. (Local gravity is g.) What checks can you make?

Answer 1

Answer:

$Ye = \frac{g*T_2^2*T_i}{2*(T_i + T_2)}$

Explanation:

Let Ye be the position of the elevator, Ve be the velocity of the elevator and Ym the position of the marble. We know that:

$Ye=Ve*T_i$

$Ym = Ye+Ve*T_2-\frac{g*T_2^2}{2}$   Since the marble hits the ground at T2, Ym=0m

Replacing Ye into the equation for Ym:

$0=Ve*T_i+Ve*T_2-\frac{g*T_2^2}{2}$   Solving for Ve:

$Ve=\frac{g*T_2^2}{2*(T_i+T_2)}$  Replcaing this value into Ye:

$Ye = \frac{g*T_2^2*T_i}{2*(T_i + T_2)}$

If we check units of Ye expression, they must be distance units.