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(C) The maximum vertical height $H_{\max}$ attained by a projectile is given by the formula $H_{\max} = \frac{v^2}{2g}$,where $v$ is the initial veloc

Vedclass · Accepted Answer

$272$

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(C) The maximum vertical height $H_{\max}$ attained by a projectile is given by the formula $H_{\max} = \frac{v^2}{2g}$,where $v$ is the initial velocity and $g$ is the acceleration due to gravity.Given $H_{\max} = 136\,m$,we have $\frac{v^2}{2g} = 136\,m$.This implies $v^2 = 272g$.The maximum horizontal range $R_{\max}$ is achieved at an angle of $45^\circ$ and is given by $R_{\max} = \frac{v^2}{g}$.Substituting the value of $v^2$ from the height equation,we get $R_{\max} = \frac{272g}{g} = 272\,m$.Alternatively,since $H_{\max} = \frac{v^2}{2g}$ and $R_{\max} = \frac{v^2}{g}$,it follows that $R_{\max} = 2H_{\max}$.Therefore,$R_{\max} = 2 	imes 136\,m = 272\,m$.

The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance up to which he can throw the same ball is $.....\,m$.

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