Answer to Question #137059 in Physics for JJ Thompson

Question

A soccer ball is dropped from a hot air balloon at a height of 1000 ft. Two seconds after the first soccer ball is dropped, a tennis ball is projected vertically upward from the ground with a velocity of 248 ft/sec. When and where will the stones pass each other?

Answer 1

Given:

t=2 s

H=1000 ft

v=248 ft/sec

Calculate at what height the soccer ball will be after t=2 seconds, this give us our reference height:

$$h=\frac{gt^2}{2}.$$

At the height of the first ball

$$y=H-h=H-\frac{gt^2}{2}$$

the tennis ball is launched and goes upward.

At some height above the ground x after time T the objects will meet. By the time T, the soccer ball will make

$$H-x=uT+\frac{gT^2}{2}=\\\space\\ =T\sqrt{2gh}+\frac{gT^2}{2}=\\\space\\ =g\bigg(tT+\frac{T^2}{2}\bigg).$$

The tennis ball will be at the same height at the same time, but its motion will be defined by another equation:

$$x=vT-\frac{gT^2}{2}.$$

Substitute for x in the above equation:

$$H-vT+\frac{gT^2}{2}=gtT+\frac{gT^2}{2},\\\space\\ T=\frac{2H-gt^2}{2(gt+v)}=2.996\text{ s},\\\space\\ x=vT-\frac{gT^2}{2}=598.5\text{ ft}.$$