Question

A ball is thrown upward from ground with an initial speed of 25 m/s, at the same instant, another ball is dropped from a building 15 meters high. a.) How long will the two balls be at the same height? b.) Calculate the height where the two balls meets.

Answer 1

Equations of motion of balls

$$y_1=25t-5t^2\\ y_2=15-5t^2$$

When two balls are at the same height

$$y_1=y_2$$

So

$$25t-5t^2=15-5t^2$$

We obtain

(a)

$$t=\frac{15}{25}=0.6\:\rm{s}$$

(b)

$$h=y_1=y_2=15-5\times 0.6^2=13.2\:\rm{m}$$