Answer to Question #160535 in Physics for Peace Simon

Question

John is on top of the building and Jack is down if John throws ball at an angle ok 60 and with initial velocity 20m/s at what height will the ball reach after 2s

Answer 1

The ball will move upward during

$$t=\frac{v\text{ sin}60°}{g}=1.77\text{ s}.$$

The height above the building top is

$$h=\frac{gt^2}{2}=15.3\text{ m}.$$

Then the ball falls during 2-1.77=0.23 s and covers a vertical distance of

$$d=\frac{gt_1^2}{2}=0.259\text{ m}.$$

Thus, the ball will be at a height of

$$H=h-d=15\text{ m}$$

above the building.