Q124 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Physics

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

  • A
    $4 \sqrt{3}$
  • B
    $\frac{\sqrt{3}}{4}$
  • Correct answer
    C
    $\frac{4}{\sqrt{3}}$
  • D
    $\frac{2}{\sqrt{3}}$

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 $\frac{\text { Range }}{\text { Max. height }}=\frac{u^2 \sin 2 \theta / g}{u^2 \sin ^2 \theta / 2 g}$ $\begin{aligned} & =\frac{2 \sin 2 \theta}{\sin ^2 \theta}=\frac{2 \times 2 \cos \theta}{\sin \theta} \\ & =4 \times \cot \theta=4 \times \cot 60^{\circ}[\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta] \\ & =4 \times \frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}\end{aligned}$

Hence, the correct answer is option: 'C'.

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