Question

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A ball is dropped from a height of 10m .At the same time another ball is thrown vertically upward at an initial speed of 10m/s .How high above the ground will the two balls collide ?

Answer 1

Step-by-step explanation:

Dropped ball height is given by

h = 10 - 1/2 at^2

Thrown ball height is given by

h = 10 t - 1/2 at^2

when are the two h's equal ?

10 - 1/2 at^2 = 10t - 1/2 at^2

10 = 10 t

t = 1 second is the time when the heights are equal:

at one second the dropped ball is

10 - ( 1/2 *9.81*1^2) = 5.095 m high

as a check: the thrown ball is

10 (1) - 1/2 *9.81 *1^2 = 5.095 m