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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2024 · Physics · Motion in Two Dimensions

A projectile is projected with velocity of 40 m / s at an angle θ...

A projectile is projected with velocity of 40m/s40 \mathrm{ m} / \mathrm{s} at an angle θ\theta with the horizontal. If RR be the horizontal range covered by the projectile and after tt seconds, its inclination with horizontal becomes zero, then the value of cot⁡θ\cot \theta is [Take, g=10m/s2g=10 \mathrm{ m} / \mathrm{s}^2 ]

Options

  1. AR20t2\frac{R}{20 t^2}
  2. BR10t2\frac{R}{10 t^2}
  3. C5Rt2\frac{5 R}{t^2}
  4. DRt2\frac{R}{t^2}

Correct answer

A. R20t2\frac{R}{20 t^2}

Step-by-step solution

At maximum height, inclination with horizontal becomes zero. Time (taken by) projectile to reach maximum height, t=usin⁡θg⇒u=gtsin⁡θ…(i)\begin{aligned} \quad t & =\frac{u \sin \theta}{g} \\ \Rightarrow \quad u & =\frac{g t}{\sin \theta}\ldots(i) \end{aligned} Since, R=ucos⁡θ×(2t)R=u \cos \theta \times(2 t)⇒cos⁡θ=R2ut⇒cos⁡θ=R2⋅gtsin⁡θ⋅t=Rsin⁡θ2gt2(Using(i))\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}⇒cot⁡θ=R2gt2=R2×10t2=R20t2\Rightarrow \quad \cot \theta=\frac{R}{2 g t^2}=\frac{R}{2 \times 10 t^2}=\frac{R}{20 t^2}
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