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 ?
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 ?
Maximum horizontal distance, $R=100 \,m$ The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is $45^{\circ},$ i.e., $\theta=45^{\circ}$ The horizontal range for a projection velocity $v$, is given by the relation
$R=\frac{u^{2} \sin 2 \theta}{g}$
$100=\frac{u^{2}}{g} \sin 90^{\circ}$
$\frac{u^{2}}{g}=100$
The ball will achieve the maximum height when it is throwertically upward. For such motion, the final velocity $v$ is zero at the maximum height $H$ Acceleration, $a=-g$
Using the third equation of motion
$v^{2}-u^{2}=-2 g H$
$H=\frac{1}{2} \times \frac{u^{2}}{g}=\frac{1}{2} \times 100=50 \,m$
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