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A projectile thrown from the ground has initial speed ' $u$ ' and its direction makes an angle ' $q$ ' with the horizontal. If at maximum height from ground, the speed of projectile is half its initial speed of projection, then the maximum height reached by the projectile is
$\begin{aligned} & \text { [g }=\text { acceleration due to gravity, } \sin 30^{\circ}=\cos 60^{\circ}=0.5, \\ & \left.\cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right]\end{aligned}$
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$\begin{aligned} & \text { [g }=\text { acceleration due to gravity, } \sin 30^{\circ}=\cos 60^{\circ}=0.5, \\ & \left.\cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right]\end{aligned}$
Solution:
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Verified Answer
The correct answer is:
$\frac{3 u^2}{8 g}$
Given at maximum height
$\begin{aligned} & \mathrm{u} \cos \theta=\frac{1}{2} \mathrm{u} \Rightarrow \cos \theta=\frac{1}{2} \\ & \therefore \theta=60^{\circ}\end{aligned}$
We know, $\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}=\frac{3 \mathrm{u}^2}{8 \mathrm{~g}}$
$\begin{aligned} & \mathrm{u} \cos \theta=\frac{1}{2} \mathrm{u} \Rightarrow \cos \theta=\frac{1}{2} \\ & \therefore \theta=60^{\circ}\end{aligned}$
We know, $\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}=\frac{3 \mathrm{u}^2}{8 \mathrm{~g}}$
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