Question

A stone is tossed upward at the moment a ball is dropped from a height of 25 meters. The stone’s initial velocity is 25 m/s. At what height will the two meet?

 

 

Answer 1

The equations of motion of the objects are given by

$$y_{\rm stone}=v_0t-gt^2/2=25t-4.9t^2$$ $$y_{\rm ball}=h-gt^2/2=25-4.9t^2$$

At the instant when objects meet

$$y_{\rm stone}=y_{\rm ball}$$

Hence

$$25t-4.9t^2=25-4.9t^2$$

$$t=1\:\rm s$$

$$y_{\rm stone}=y_{\rm ball}=25-4.9*1^2=20.1\:\rm m$$