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.