Question

A boy and his sister stand on a balcony 14.0 m above the ground. The boy throws a ball velocity downward with an initial speed of 4.00 m/s. At the same instant, his sister throws a ball velocity upward with the same initial speed. When the first ball hits the ground, what is the vertical separation between the two balls.

Answer 1

The law of motion of the "boy's ball"

$$y_1=14.0-4.00t-\frac{9.81t^2}{2}$$

The law of motion of the "girl's ball"

$$y_2=14.0+4.00t-\frac{9.81t^2}{2}$$

The time of motion of the first ball is given by equation

$$14.0-4.00t-\frac{9.81t^2}{2}=0$$ $$t=1.33\:\rm s$$

The position of the second ball at this instant

$$y_2=14.0+4.00\times 1.33-\frac{9.81\times (1.33)^2}{2}=10.6\: \rm m$$

Answer: $10.6\:\rm m$