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Definition of a Projectile

A projectile ...

Question

A projectile thrown with an initial speed $v_{0}$ has the same range $R$ when the maximum height attained by it is either $h_{1}$ or $h_{2} .$ Then $R, h_{1}$ and $h_{2}$ are related as

$R = \sqrt{h_{1} h_{2}}$

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$R = 2 \sqrt{h_{1} h_{2}}$

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$R = 3 \sqrt{h_{1} h_{2}}$

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$R = 4 \sqrt{h_{1} h_{2}}$

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Solution

The correct option is D
$R = 4 \sqrt{h_{1} h_{2}}$

The range of a projectile is the same for two angles of projection $θ$ and $90^{\circ} - θ .$ For these two angles of projection, the maximum heights are

$h_{1} = \frac{v_{0}^{2} {sin}^{2} θ}{2 g}$
and

$h_{2} = \frac{v_{0}^{2} {sin}^{2} (90^{\circ} - θ)}{2 g} = \frac{v_{0}^{2} {cos}^{2} θ}{2 g}$

$∴ h_{1} h_{2} = \frac{v_{0}^{4} {sin}^{2} θ {cos}^{2} θ}{4 g^{2}}$

Also, $R = \frac{2 v_{0}^{2} sin θ cos θ}{g}$

$R^{2} = \frac{4 v_{0}^{4} {sin}^{2} θ {cos}^{2} θ}{g^{2}} = 16 \times \frac{v_{0}^{4} {sin}^{2} θ {cos}^{2} θ}{4 g^{2}}$

$R^{2} = 16 h_{1} h_{2}$

$R = 4 \sqrt{h_{1} h_{2}}$

Hence, the correct choice is $(D)$ .

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