Calculation of Same Height Time for Thrown Ball and Dropped Ball

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

$${h_1(t)=h_2(t)}$$

$${25t-\frac{1}{2}(10)t^2}={15-\frac{1}{2}(10)t^2}$$

$${25t=15}$$

$${t=\frac{15}{25}=0.6 s}$$