Question

At what angle of elevation , should a projectile with velocity 20 ms-1 so as to reach a maximum height of 10 m :

Answer 1

Hey There,

Here, initial velocity is u = 20 m/s
and Maximum Height is H = 10 m

The formula for maximum height in Projectile Motion is given as : 
$H=\frac{u^2sin^2\theta}{2g} \\ \\ \implies 10=\frac{20^2sin^2\theta}{2\times10} \\ \\ \implies 10 = 20 sin^2\theta \\ \\ \implies sin^2\theta=\frac{1}{2} \\ \\ \implies sin \theta=\frac{1}{\sqrt{2}} \\ \\ \implies \boxed{\theta=45^{\circ}}$


Hope this helps
Purva
Brainly Community

Answer 2

The angle is: 45°

Explanation:

In this,H is given as 10m.

u as 20m/s.

Then using H formula the angle is calculated.