Question:
A ball is thrown upward from the ground with an initial speed of 25 ms-1 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?
Solution:
Thrown up case
Initial velocity of the ball u=25 ms -1
Let y1 be the height from the ground
$${y=ut-\frac{1}{2}gt^2}$$
$${y=25t-\frac{1}{2}(10)t^2}$$
Let it be h1(t)
In the case of the ball when dropped
Let (15-y) be the displacement from the top
$${y = 15-\frac{1}{2}(10)t^2)}$$
Let it be h2(t)
The balls are at the same height
$${h_1(t)=h_2(t)}$$
$${25t-\frac{1}{2}(10)t^2}={15-\frac{1}{2}(10)t^2}$$
$${25t=15}$$
$${t=\frac{15}{25}=0.6 s}$$