Question

John is on top of the building and Jack is down. If John throws ball at an angle of 60° and with initial velocity 20m/s at what height will the ball reach after 2s? Consider the motion along the y direction. Show your solution. (The height of the building was not given)

Answer 1

The ball will move upward during

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

The height above the building top is

$$h=0.5gt^2=15.3\ m$$

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

$$d=0.5gt^2_1=0.3\ m$$

Thus, the ball will be at a height of

$$H=h-d=15 \ m$$

above the building.