AP EAPCET 07-Jul-2022 Shift 2 Paper

Given situation is shown below :

Here we are considering only vertical motion of ball.
Given, initial vertical velocity of ball
uy=u‌sin‌θ=20×sin‌30∘
=20×

1

2

=10m∕ s
Vertical acceleration,
ay=g=−10m∕ s2
Time to reach the ground
=3 s

Now using,
h=uyt+

1

2

ayt2
We get,
h=10×3+

1

2

(−10)×32
⇒h=30−45=−15m
So height of building is 15m. (negative sign shows downward displacement of ball).