For a projectile, if alpha is the angle of pr | vedclass

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(C) The maximum height $h$ is given by $h = \frac{u^2 \sin^2 \alpha}{2g}$.The range $R$ is given by $R = \frac{u^2 \sin 2\alpha}{g} = \frac{2 u^2 \sin

Vedclass · Accepted Answer

$	an \alpha=\frac{4 h}{R}, h=\frac{g T^2}{8}$

Vedclass · Answer

(C) The maximum height $h$ is given by $h = \frac{u^2 \sin^2 \alpha}{2g}$.The range $R$ is given by $R = \frac{u^2 \sin 2\alpha}{g} = \frac{2 u^2 \sin \alpha \cos \alpha}{g}$.Dividing $h$ by $R$:$\frac{h}{R} = \frac{u^2 \sin^2 \alpha}{2g} \cdot \frac{g}{2 u^2 \sin \alpha \cos \alpha} = \frac{	an \alpha}{4}$.Therefore, $	an \alpha = \frac{4h}{R}$.The time of flight $T$ is given by $T = \frac{2u \sin \alpha}{g}$.Squaring both sides, $T^2 = \frac{4u^2 \sin^2 \alpha}{g^2}$.Thus, $\frac{g T^2}{8} = \frac{g}{8} \cdot \frac{4u^2 \sin^2 \alpha}{g^2} = \frac{u^2 \sin^2 \alpha}{2g} = h$.So, $h = \frac{g T^2}{8}$.

For a projectile, if $\alpha$ is the angle of projection, $R$ is the range, $h$ is the maximum height, and $T$ is the time of flight, then:

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