A cricketer can throw a ball to a maximum hor | vedclass

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(B) The maximum horizontal range $R$ is given by $R = \frac{u^2}{g}$,where $u$ is the initial velocity and $g$ is the acceleration due to gravity. Giv

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) The maximum horizontal range $R$ is given by $R = \frac{u^2}{g}$,where $u$ is the initial velocity and $g$ is the acceleration due to gravity. Given $R = 100 \; m$,we have $\frac{u^2}{g} = 100 \; m$.To achieve the maximum vertical height $H$,the ball must be thrown vertically upward at an angle of $90^{\circ}$.The formula for maximum vertical height is $H = \frac{u^2}{2g}$.Substituting the value of $\frac{u^2}{g} = 100 \; m$ into the height formula:$H = \frac{1}{2} 	imes \left( \frac{u^2}{g} ight) = \frac{1}{2} 	imes 100 \; m = 50 \; m$.Thus,the cricketer can throw the ball to a maximum height of $50 \; m$.

$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \; m$. How much high above (in $m$) the ground can the cricketer throw the same ball?

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