Question

the speed of a projectile when it is at its greatest height is square root of 2/5 times its speed at half the maximum height the angle of projection is

Answer 1

Theta = 60 deg.

Note that velocity is same at the same height on the ascent or descent.
See the solution enclosed.

Answer 2

The angle of projection is 60°.

Solution:

Let us consider that the maximum height covered by the projectile is H and the angle of projection is θ.

It is given that,

$v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}$

So the maximum height of a projectile motion can be found as

$H=\frac{u^{2} \sin ^{2} \theta}{2 g} \quad \rightarrow(1)$

In the above equation, u is the initial velocity and g is the acceleration due to gravity.

We know that velocity at maximum height is $v_{H}=u \cos \theta$

Squaring on both sides, we get $v_{H}^{2}=u^{2} \cos ^{2} \theta \quad \rightarrow(2)$

From Newton’s second law of motion:

$v^{2}=u^{2}-2 a s \quad \rightarrow(3)$

We can substitute $v=v_{H / 2}$, as the velocity at half the maximum height and displacement (s) can be replaced by half the maximum height, i.e., $\mathrm{s}=\mathrm{H} / 2\ in\ (3)$

$v_{H / 2}^{2}=u^{2}-2 g\left(\frac{H}{2}\right) \quad \rightarrow(4)$

The H in eqn (4) can be replaced with the value of H in eqn (1)

$v_{H / 2}^{2}=u^{2}-2 g\left(\frac{\frac{u^{2} \sin ^{2} \theta}{2 g}}{2}\right) \quad \rightarrow(5)$

$v_{H / 2}^{2}=u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right) \quad \rightarrow(6)$

It is given that $v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}$  

So squaring on both sides, we get  

$v_{H}^{2}=\frac{2}{5} v_{H / 2}^{2} \rightarrow(7)$

So, substitute eqn (2) and eqn (6) in eqn (7)

$u^{2} \cos ^{2} \theta=\frac{2}{5}\left[u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right)\right]$

(We know that, $\cos ^{2} \theta+\sin ^{2} \theta=1$)

$\therefore u^{2}\left(1-\sin ^{2} \theta\right)=\frac{2 u^{2}}{5}\left[1-2\left(\frac{\sin ^{2} \theta}{4}\right)\right]$

$\left(1-\sin ^{2} \theta\right)=\frac{2}{5}\left[1-\left(\frac{\sin ^{2} \theta}{2}\right)\right]$

$1-\sin ^{2} \theta=\frac{2}{5}-\frac{1}{5} \sin ^{2} \theta$

$\frac{1}{5} \sin ^{2} \theta-\sin ^{2} \theta=\frac{2}{5}-1$

$-\frac{4}{5} \sin ^{2} \theta=-\frac{3}{5}$

$\sin ^{2} \theta=\frac{3}{5} \times \frac{5}{4}$

$\sin ^{2} \theta=\frac{3}{4}$

$\therefore \sin \theta=\frac{\sqrt{3}}{2}$

$\bold{\therefore \theta=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^{\circ}}$

Thus the angle of projection is 60°.