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

`Ye = (g*T_2^2*T_i)/(2*(T_i + T_2))`

Step-by-step 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-(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-(g*T_2^2)/(2)` Solving for Ve:

`Ve=(g*T_2^2)/(2*(T_i+T_2))` Replcaing this value into Ye:

`Ye = (g*T_2^2*T_i)/(2*(T_i + T_2))`

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