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Question

(C) The maximum height $H$ reached by a projectile is given by the formula: $H = \frac{u^2 \sin^2 	heta}{2g}$,where $u$ is the initial velocity and $

Vedclass · Accepted Answer

$2 \sqrt{\frac{2 H}{g}}$

Vedclass · Answer

(C) The maximum height $H$ reached by a projectile is given by the formula: $H = \frac{u^2 \sin^2 	heta}{2g}$,where $u$ is the initial velocity and $	heta$ is the angle of projection.From this,we can find the vertical component of the initial velocity,$u_y = u \sin 	heta = \sqrt{2gH}$.The time of flight $T$ for a projectile is given by the formula: $T = \frac{2u \sin 	heta}{g}$.Substituting the value of $u \sin 	heta$ into the time of flight formula:$T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2gH}{g^2}} = 2 \sqrt{\frac{2H}{g}}$.Therefore,the correct option is $C$.

$A$ body when projected at an angle $\theta$ with the horizontal reaches a maximum height $H$. The time of flight of the body will be ($g=$ acceleration due to gravity).

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