A ball s projected from a point on horizontal ground. The speed of projection is 30 m/s and the greatest height reached is 20 meter. Assuming no air resistance, find the angle of projection above the horizontal ?

1 Answer
Jun 28, 2017

The angle is =41.3º

Explanation:

Resolving in the vertical direction uarr^+

The initial velocity is u=30sintheta

The height is h=s=20m

The acceleration is a=(-g)ms^-2

The velocity, at the greatest height is v=0ms^-1

We apply, the equation of motion

v^2=u^2+2as

0=(30sintheta)^2-2*g*20

(30sintheta)^2=40g

30sintheta=sqrt(40g)

sintheta=sqrt(40g)/30=0.66

theta=41.3º