Q91 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics

A projectile is projected with velocity of $40 \mathrm{~m} / \mathrm{s}$ at an angle $\theta$ with the horizontal. If $R$ be the horizontal range covered by the projectile and after $t$ seconds, its inclination with horizontal becomes zero, then the value of $\cot \theta$ is [Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

  • Correct answer
    A
    $\frac{R}{20 t^2}$
  • B
    $\frac{R}{10 t^2}$
  • C
    $\frac{5 R}{t^2}$
  • D
    $\frac{R}{t^2}$

Solution

At maximum height, inclination with horizontal becomes zero. Time (taken by) projectile to reach maximum height, $\begin{aligned} \quad t & =\frac{u \sin \theta}{g} \\ \Rightarrow \quad u & =\frac{g t}{\sin \theta}\ldots(i) \end{aligned}$ Since, $R=u \cos \theta \times(2 t)$ $\begin{aligned} \Rightarrow \quad \cos \theta & =\frac{R}{2 u t} \\ \Rightarrow \quad \cos \theta & =\frac{R}{2 \cdot \frac{g t}{\sin \theta} \cdot t} \\ & =\frac{R \sin \theta}{2 g t^2}(Using(i)) \end{aligned}$ $\Rightarrow \quad \cot \theta=\frac{R}{2 g t^2}=\frac{R}{2 \times 10 t^2}=\frac{R}{20 t^2}$

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

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