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# A person can throw a stone to a maximum distance of 100 m. The greatest height to which he can throw the stone is:-(A) 100m(B) 75m(C) 50m(D) 25m

Last updated date: 16th Sep 2024
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Hint: The maximum distance given, 100m, is the range of projectile motion. The maximum height is achieved when thrown straight up at ${45^ \circ }$ angle. Thus launching angle is $\theta = {45^ \circ }$.
Formula Used: The formulae used in the solution are given here.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta$ is the launch angle and $g$ is the acceleration due to gravity and the maximum height is given by $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$

Complete Step by Step Solution: This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta$ is the launch angle and $g$ is the acceleration due to gravity.
Since, $R = 100m$, thus $\dfrac{{{u^2}\sin 2\theta }}{g} = 100$ and the acceleration due to gravity $g = 9.8$.
Thus, it can be written that, ${u^2}\sin 2\theta = 100g$ where $g = 9.8$
It is known to us that, the greatest range is achieved using a ${45^ \circ }$ launch angle:
${u^2}\sin \left( {2 \times {{45}^ \circ }} \right) = 980$
$\Rightarrow {u^2}\sin {90^ \circ } = 980$
We know that, $\sin {90^ \circ } = 1$,
$\therefore {u^2} = 980$
The height is achieved when thrown straight up at ${45^ \circ }$ angle.
Thus, height $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.
Since, ${u^2} = 980$, $\theta = {45^ \circ }$ and $g = 9.8$
$H = \dfrac{{980}}{{39.2}}$
$\Rightarrow H = 25m$

The greatest height to which he can throw the stone is 25m. Hence, the correct answer is Option D.