A projectile is thrown with some initial velo | vedclass

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Question

(c)

$(u \cos \alpha)=\sqrt{\frac{2}{5}} \sqrt{\left(u \cos \alpha^2\right)+\left\{(u \sin \alpha)^2-2 g h\right\}}$

Here, $h=\frac{H}{2}=\frac{u

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(c)

$(u \cos \alpha)=\sqrt{\frac{2}{5}} \sqrt{\left(u \cos \alpha^2\right)+\left\{(u \sin \alpha)^2-2 g h\right\}}$

Here, $h=\frac{H}{2}=\frac{u^2 \sin ^2 \alpha}{4 g}$

Solving this equation, we get $\alpha=60^{\circ}$

A projectile is thrown with some initial velocity at an angle $\alpha$ to the horizontal. Its velocity when it is at the highest point is $(2 / 5)^{1 / 2}$ times the velocity when it is at height half of the maximum height. Find the angle of projection $\alpha$ with the horizontal.

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