A tennis ball is shot straight up from the ground. After traveling 4.00 m, the ball passes a 4.00 m high window and then continues straight up until it loses all its upward velocity and falls to the ground. During the last second of its flight before it hits the ground, the ball drops 20.0 m. a) What is the maximum height above the ground reached by the ball?
b) What is the total time for the ball’s flight?
c) When the ball is travelling upward, during what time interval is the ball passing the window?

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{20-\frac{1}{2}\times 9.81\times 1^2}{1}\\\Rightarrow u=15.095\ m/s[/tex]

At the beginning of the last 1 second the velocity will be 15.095 m/s

Taking that as the final velocity of the path from the maximum height we find time

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{15.095-0}{9.81}\\\Rightarrow t=1.539\ s[/tex]

So, the ball traveled for 1+1.539 = 2.2539 seconds from the maximum height to the ground.

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 2.2539^2\\\Rightarrow s=24.918\ m[/tex]

The maximum height the ball traveled is 24.918 m.

The time the ball will take to go up is 2.2539 seconds

So, total time the ball will be in flight is 2.2539+2.2539 = 4.5078 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{24.918-\frac{1}{2}\times -9.81\times 2.2539^2}{2.2539}\\\Rightarrow u=22.11\ m/s[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 4=22.11t+\frac{1}{2}\times -9.81\times t^2\\\Rightarrow 4.905t^2-22.11t+4=0[/tex]

Solving the above equation we get

t = 0.18 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 8=22.11t+\frac{1}{2}\times -9.81\times t^2\\\Rightarrow 4.905t^2-22.11t+8=0[/tex]

Solving the equation we get

t = 0.39 seconds

So, the ball will traveling in the [0.18, 0.39] seconds of the flight