Answer to Question #131590 in Physics for Alba Isturiz

Question

A ball is dropped from rest at the top of a 100 m tall building. At the same instant, a second ball is thrown upward from the base of the building with an initial velocity of 50 m/s. When will the two balls collide, and at what distance above the street? 

Answer 1

The law of motion of the first ball

$y_1=y_0+v_{0y}t-\frac{gt^2}{2}\\ =100-\frac{9.8t^2}{2}=100-4.9t^2$

The law of motion of the second ball

$y_2=y_0+v_{0y}t-\frac{gt^2}{2}\\ =50t-\frac{9.8t^2}{2}=50t-4.9t^2$

At the instant when balls collide $y_1=y_2,$ so

$100-4.9t^2=50t-4.9t^2$ $t=2\:\rm s$

The distance above the street

$d=y_1(t)=y_2(t)=50\times 2-4.9\times 2^2=80.4\:\rm m$