A cricketer can throw a ball to maximum horiz | vedclass

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Question

Let $u$ be the velocity of projection of the ball.

The ball will cover maximum horizontal

distance when angle of projection with

horizontal,

Vedclass · Accepted Answer

$50$

Vedclass · Answer

Let $u$ be the velocity of projection of the ball.

The ball will cover maximum horizontal

distance when angle of projection with

horizontal, $	heta=45^{\circ} .$ Then

$\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}$

Here, $\mathrm{R}_{\max }=100 \mathrm{m}$

$	herefore \quad \frac{\mathrm{u}^{2}}{\mathrm{g}}=100 \mathrm{m}$

As $v^{2}-u^{2}=2 a s$

Here. $\mathrm{v}=0$ (At highest point velocity is zeno)

$a=-g, s=H$

$	herefore H=\frac{u^{2}}{2 g}=\frac{100}{2}=50 \mathrm{m} \quad(\mathrm{Using}(\mathrm{i}))$

A cricketer can throw a ball to maximum horizontal distance of $100\,m$ . With the same speed how much high above the ground can the cricketer throw the same ball ......... $m$

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