Answer to Question #350199 in Physics for slv

Question

There’s two rocks. On of them is at the height of 125m from earth and is going to start falling. The other was thrown up with a speed of 50 meters per second. When at what height will they meet? (Sorry if there are mistakes English isn’t my native language)

Answer 1

The equations of motion of rocks are given by

$$y_1(t)=y_{01}+v_{01}t-gt^2/2\\ y_1(t)=125-4.9t^2$$

$$y_2(t)=y_{02}+v_{02}t-gt^2/2\\ y_2(t)=50t-4.9t^2$$

At the meeting point

$$y_1(t)=y_2(t)$$

$$125-4.9t^2=50t-4.9t^2$$

$$t=125/50=2.5\:\rm s$$

The corresponding height

$$y=125-4.9*2.5^2=94.4\:\rm m$$