The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is :
Solution
For vertical throw,
<br/><br/>
$$
\begin{aligned}
& h=\frac{v^{2}}{2 g} \\\\
& v=\sqrt{2 g h}=\sqrt{2 g \times 136} \quad...(1)
\end{aligned}
$$
<br/><br/>
For max range, $\theta=45^{\circ}$
<br/><br/>
$R_{\max }=\frac{v^{2}}{g} \quad...(2)$
<br/><br/>
From (1) and (2)
<br/><br/>
$$
\begin{aligned}
R_{\max } & =\frac{v^{2}}{g}=\frac{2 g \times 136}{g} \\\\
& =272 \mathrm{~m}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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