Question

Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.24 s as it moves a distance of 1.18 m from the bottom to the top of the window.How long does it take before the ball reappears?What is the greatest height of the ball above the top of the window?

Answer 1

Using the second equation of motion which is,

y=ut+0.5at²

Where y is the displacement, a is the acceleration and t is the time taken

and u is the initial velocity, which is not given in this problem

So,

1.18=u(0.24)+[0.5*-9.81*0.24²]

u=6.094 m/s

To determine when the ball reappears at the top of the window, use V = Vo + at

0=6.094-9.81t

t=0.62 seconds

It takes 0.62s to drop from the trajectory peak (zero motion) to the bottom of the window. Subtract the travel time past the window 0.24s to find the travel time from peak to the top of the window.
0.62 - 0.24 = 0.35 s

So, from the time the ball passes the top of the window on the way up, to when it passes on the way down is
2 * 0.38

= 0.76 s

To find the height of the ball above the top of the window use

V² = u² + 2ay
We know everything except S, so plug in the known variables and solve for S, which will be the distance from window bottom to peak trajectory, which is where V = 0.

0 = 6.094² + 2(-9.81)S

S = 1.89 m

Subtrack the length of the window 1.18 from 1.89

= 0.71 m the height of the ball above the top of the window.