In the projectile motion of an object, the object reaches its maximum height where its speed is half of…
In the projectile motion of an object, the object reaches its maximum height where its speed is half of initial speed. Then the ratio between range and maximum height of projectile is
$4 \sqrt{3}$
$\frac{\sqrt{3}}{4}$
$\frac{4}{\sqrt{3}}$
$\frac{2}{\sqrt{3}}$
Solution
At maximum height, speed of projectile will be $u \cos \theta$
Now, given $u \cos \theta=\frac{1}{2} u \Rightarrow \cos \theta=\frac{1}{2}$ or $\theta=60^{\circ}$
Now, ratio of range and maximum height of projectile is $\frac{\text { Range }}{\text { Max. height }}=\frac{u^2 \sin 2 \theta / g}{u^2 \sin ^2 \theta / 2 g}$
$\begin{aligned} & =\frac{2 \sin 2 \theta}{\sin ^2 \theta}=\frac{2 \times 2 \cos \theta}{\sin \theta} \\ & =4 \times \cot \theta=4 \times \cot 60^{\circ}[\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta] \\ & =4 \times \frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}\end{aligned}$