Question

A ball is thrown from ground level so that it just
clears a wall 3 m high when it is moving
horizontally. If the initial speed of the ball is 20 m/
s, find the angle of projection. [g=10m/s?)​

Answer 1

Answer:

The value of angle of projectile would be $22.7^\circ$.

To find:

The angle of projection $(\theta)$

Solution:

Since the ball is passing 3 metre then its maximum height is three meter.

Let us assume that the angle of projection be $\theta$.

The equation of maximum height$\frac{(v^2sin^2 \theta)}{2g}$ is the equation that has to be applied now.  

H = $\frac{(v^2sin^2\theta)}{2g}$

Applying values in the above equation

3= \frac{(20^2 times sin^2\theta)}{2\times10}

Sin$\theta$= $\sqrt{\frac{3}{20}}$

Sin$\theta$= 0.387  

$\theta$ = $22.7^\circ$

Result:

The angle of projection would be $22.7^\circ$.