Question

A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 56.7 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to

Answer 1

Calculate the time it will take to go upward until the ball stops:

$$t=v/g.$$

Calculate the height above these 60 m that the ball reaches:

$$h=\frac{v^2}{2g}.$$

Then, from this height plus 60 m, the ball falls, and it takes a time

$$\tau=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2(h+60)}{g}}=\sqrt{\frac{2(60+\frac{v^2}{2g})}{g}}.$$

The total time is time upward plus time downward:

$$T=t+\tau=\frac{v}{g}+\sqrt{\frac{2(60+\frac{v^2}{2g})}{g}},\\\space\\ T=\frac{56.7}{9.8}+\sqrt{\frac{2(60+\frac{56.7^2}{2\cdot9.8})}{9.8}}=12.5\text{ s}.$$