Physics - Kinematics Question with Solution | TestHub

Solution:

Let $u$ be the initial velocity and $\alpha$ be the angle of projection.

The range $R$ is given by $R=\frac{u^2\sin (2\alpha )}{g}$.

The maximum height $h$ is given by $h=\frac{u^2{\sin }^2\alpha }{2g}$.

For a given range $R$, there are two possible angles of projection, $\alpha$ and $(90^{\circ }-\alpha )$, provided $R\le \frac{u^2}{g}$.

Let $h_1$ and $h_2$ be the maximum heights for these two angles.

$h_1=\frac{u^2{\sin }^2\alpha }{2g}$

$h_2=\frac{u^2{\sin }^2(90^{\circ }-\alpha )}{2g}=\frac{u^2{\cos }^2\alpha }{2g}$

The product of the two heights is:

$h_1{h}_2=\left(\frac{u^2{\sin }^2\alpha }{2g}\right)\left(\frac{u^2{\cos }^2\alpha }{2g}\right)=\frac{u^4{\sin }^2\alpha {\cos }^2\alpha }{4g^2}$

$h_1{h}_2=\frac{u^4(2\sin \alpha \cos \alpha {)}^2}{16g^2}=\frac{u^4{\sin }^2(2\alpha )}{16g^2}$

We know that $R=\frac{u^2\sin (2\alpha )}{g}$, so $R^2=\frac{u^4{\sin }^2(2\alpha )}{g^2}$.

Substituting this into the expression for $h_1{h}_2$:

$h_1{h}_2=\frac{R^2}{16}$

$R^2=16h_1{h}_2$

$R=\sqrt{16h_1{h}_2}=4\sqrt{h_1{h}_2}$

The problem uses $h$ and $h^{\prime }$ for the two greatest heights, so $R=4\sqrt{h{h}^{\prime }}$.

The final answer is $\boxed{R=4\sqrt{h{h}^{\prime }}}$.