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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Motion in Two Dimensions

JIPMER 2016 · Physics · Motion in Two Dimensions

A stone projected with a velocity u at an angle θ with the...

A stone projected with a velocity uu at an angle θ\theta with the horizontal reaches maximum height H1H_1. When it is projected with velocity uu at an angle (π2−θ)\left(\frac{\pi}{2}-\theta\right) with the horizontal, it reaches maximum height H2H_2. The relation between the horizontal range RR of the projectile, H1andH2H_1 and H_2 is

Options

  1. AR=4H1H2R=4 \sqrt{H_1 H_2}
  2. BR=4(H1−H2)R=4\left(H_1-H_2\right)
  3. CR=4(H1+H2)R=4\left(H_1+H_2\right)
  4. DR=H12H22R=\frac{H_1^2}{H_2^2}

Correct answer

A. R=4H1H2R=4 \sqrt{H_1 H_2}

Step-by-step solution

We know that, H1=u2sin⁡2θ2gH_1=\frac{u^2 \sin ^2 \theta}{2 g} and  H2=u2sin⁡2(90−θ)2g⇒H2=u2cos⁡2θ2g∵H1H2=u2sin⁡2θ2g×u2cos⁡2θ2g\begin{aligned} & \text{\text{ and } } \quad H_2=\frac{u^2 \sin ^2(90-\theta)}{2 g} \\ & \Rightarrow \quad H_2=\frac{u^2 \cos ^2 \theta}{2 g} \\ & \because \quad H_1 H_2=\frac{u^2 \sin ^2 \theta}{2 g} \times \frac{u^2 \cos ^2 \theta}{2 g}\end{aligned}=\frac{4}{4} \times \frac{U^2 \sin ^2 \theta}{2 g} \times \frac{U^2 \cos ^2 \theta}{2 g} =\frac{\left(u^2 \sin 2\theta\right)^2}{16 g^2} \quad\left[\because \sin ^2 \theta=2 \sin \theta \cos \theta\right)]=R216R2=6H1H2∴R=4H1H2\begin{aligned} & =\frac{R^2}{16} \\ R^2 & =6 H_1 H_2 \\ \therefore \quad R & =4 \sqrt{H_1 H_2}\end{aligned}$$
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