Question

A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height above the ground?

Answer 1

Answer:

t = o.6 s

Explanation:

Let ball thrown from below be A and ball dropped from above be B.

A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know

d + h = 15 m  ---------------(1)

Now apply s = ut + $\frac{1}{2}$at²

To A upwards,

d = 25t - $\frac{1}{2}$gt²  -----------------(2)

To B downwards,

h = 0 + $\frac{1}{2}$gt² ----------------(3)

(1) = (2) + (3) ⇒ 15 = 25t

                         t = 0.6 s