How long does it take for the ball to reappera? what is the max height of the ball above the top of the window

Question

A ball is moving straight upward outside a window. The ball is visible for 0.25 seconds as it moves a distance of 1.05 meters from the bottom to the top of the window.

Answer 1

The equation of motion of the ball is h = V0*t - 0.5*g*t^2.  Let V0 be the ball velocity at the bottom of the window, then it moves a distance 1.05 m in 0.25 sec so

1.05 = V0*0.25 - 0.5*9.8*(0.25^2)

1.05 + .30625 = 0.25*V0

V0 = 5.425 m/s

The velocity at the top of the window is V0 - g*0.25 = 2.975 m/s.  The time to reach the top of trajectory from that point is 2.975 - g*tm = 0, tm = 0.3036 s  It takes the same time to come down to that point, so the total time it takes is twice this or 0.607 s.  The max height is hm = 2.975*tm - 0.5*g*tm^2 = 0.5*2.975^2 / g = 0.452 m above the top of the window.