Question

Sitting in a second story apartment a physicist notices a ball moving upwards just outside her window the ball is visible for 0.25 seconds as it moves a distance from the bottom to the top of the window. How much time does it take for the ball to reappear

Answer 1

The equation for the height h(t) of an object in free fall at time t with initial velocity v₀ and initial height h₀ is given by:

`h(t)=h_0 + v_0t -(g)/(2)t^2`

The change of the height(height of the window) h with time t=0.25s:

`h=v_0t-(g)/(2)t^2`

Solving for v₀, the velocity of the ball at the bottom of the window:

(i)
`v_0=(h)/(t) +(g)/(2)t`

The equation for the velocity of an object in free fall with time T:

`v=v_0-gT`

The moment the ball reaches the top of it's flight the velocity v=0:

(ii)
`v_0=gT`

Setting equations (i) and (ii) equal and solving for T:

`T=(h)/(gt)+(t)/(2)`

The time U it takes the ball to reappear is the difference between the flight time T and twice the time t it takes to cross the window.

`U= T-2t=(h)/(gt)+(t)/(2)-2t=(h)/(gt)-(3)/(2)t=(4h)/(g)-(3)/(8)`