AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, height of tower, h=19.6m Angle of line joining from point of projection to the point of ground with horizontal,
θ = 45°
In Δ‌AOB,tan‌θ=

h

R

tan‌45°=

h

R

⇒1=

h

R

⇒R=h=19.6m
where, R be the horizontal distance.
We know that, time taken to reach from point A to B,
t=√

2h

g

By substituting the values, we get
t=√

2×19.6

9.8

=2s
Now,
Horizontal distance = Horizontal velocity × Time
⇒ R = ut
19.6 = u × 2 [By substituting the values of R and t, we get]
∴ Initial velocity, u=9.8m∕s