If the maximum height and range of a projectile are $3 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively, then the velocity of the projectile is (Take, $g=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

Here, maximum height, $H=3 \mathrm{~m}$ and range, $R=4 \mathrm{~m}$ As, maximum height, $H=\frac{u^2 \sin ^2 \theta}{2 g}$ and maximum range, $R=\frac{u^2 \sin 2 \theta}{g}$

The ratio, $ \begin{array}{rlrl} & & \frac{H}{R} & =\frac{3}{4}=\frac{\sin ^2 \theta}{2 \sin 2 \theta} \Rightarrow 3 \sin 2 \theta=2 \sin ^2 \theta \\ \Rightarrow \quad & \tan \theta & =3 \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta) \end{array} $

From Eqs. (i) and (ii), we get $ \begin{aligned} & 3 \times 2 \times 10=u^2 \sin ^2 \theta \Rightarrow 3 \times 2 \times 10=u^2 \times\left(\frac{3}{\sqrt{10}}\right)^2 \\ & u^2=\left(\frac{\sqrt{10}}{3}\right)^2 \times 3 \times 2 \times 10 \Rightarrow u^2=\frac{100 \times 2}{3} \Rightarrow u=10 \sqrt{\frac{2}{3}} \end{aligned} $ Hence, the correct option is (c)

If the maximum height and range of a projectile are $3 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively, then…

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