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

Answer:

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:

$h(t)=h_0 + v_0t -\frac{g}{2}t^2$

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

$h=v_0t-\frac{g}{2}t^2$

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

(i) $v_0=\frac{h}{t} +\frac{g}{2}t$

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:

$T=\frac{h}{gt}+\frac{t}{2}$

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.

$U= T-2t=\frac{h}{gt}+\frac{t}{2}-2t=\frac{h}{gt}-\frac{3}{2}t=\frac{4h}{g}-\frac{3}{8}$