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3-2.Motion in Plane
medium
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 ......... $m$
A
$50$
B
$100$
C
$150$
D
$200$
Solution
Let $u$ be the velocity of projection of the ball.
The ball will cover maximum horizontal
distance when angle of projection with
horizontal, $\theta=45^{\circ} .$ Then
$\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}$
Here, $\mathrm{R}_{\max }=100 \mathrm{m}$
$\therefore \quad \frac{\mathrm{u}^{2}}{\mathrm{g}}=100 \mathrm{m}$
As $v^{2}-u^{2}=2 a s$
Here. $\mathrm{v}=0$ (At highest point velocity is zeno)
$a=-g, s=H$
$\therefore H=\frac{u^{2}}{2 g}=\frac{100}{2}=50 \mathrm{m} \quad(\mathrm{Using}(\mathrm{i}))$
Standard 11
Physics
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