A stone projected with a velocity a at an angle θ with the horizontal reaches maximum height H_ 1 . When it is projected with velocity u at an angle ((π)/(2)-θ) with the horizontal, it reaches maximum height H_ 2 . The

Correct answer: A. R=4 H_ 1 H_ 2 aligned & H_ 1 = u^ 2 ^ 2 θ 2 g and H_ 2 = u^ 2 ^ 2 (90^ ^ -θ) 2 g = u^ 2 ^ 2 θ 2 g H_ 1 H_ 2 =& u^ 2 ^ 2 θ 2 g × u^ 2 ^ 2 θ 2 g = (u^ 2 2 θ)^ 2 16 g^ 2 = R^ 2 16 R=4 H_ 1 H_ 2 aligned

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Motion in Two Dimensions

BITSAT 2016 · Physics · Motion in Two Dimensions

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

A stone projected with a velocity a at an angle

\theta

with the horizontal reaches maximum height

\mathrm{H}_{1}

. When it is projected with velocity u at an angle

\left(\frac{\pi}{2}-\theta\right)

with the horizontal, it reaches maximum height

\mathrm{H}_{2}

. The relation between the horizontal range

\mathrm{R}

of the projectile, heights

\mathrm{H}_{1} and \mathrm{H}_{2}

Options

A $\mathrm{R}=4 \sqrt{\mathrm{H}_{1} \mathrm{H}_{2}}$
B $\mathrm{R}= 4 \left(\mathrm{H}_{1}-\mathrm{H}_{2}\right)$
C $\mathrm{R}= 4 \left(\mathrm{H}_{1}+\mathrm{H}_{2}\right)$
D $\mathrm{R}=\frac{\mathrm{H}_{1}^{2}}{\mathrm{H}_{2}^{2}}$

Correct answer

A. $\mathrm{R}=4 \sqrt{\mathrm{H}_{1} \mathrm{H}_{2}}$

Step-by-step solution

\begin{aligned} & \mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{ g}} \\ \text{\text{ and } } \mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{{}^{\circ}}-\theta\right)}{2 \mathrm{ g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{ g}} \\ \mathrm{H}_{1} \mathrm{H}_{2}=& \frac{u^{2} \sin ^{2} \theta}{2 \mathrm{ g}} \times \frac{u^{2} \cos ^{2} \theta}{2 \mathrm{ g}}=\frac{\left(u^{2} \sin 2 \theta\right)^{2}}{16 g^{2}}=\frac{R^{2}}{16} \\ \therefore R=4 \sqrt{H_{1} H_{2}} \end{aligned}

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