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
Answer
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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.
Note: Some assumptions have to be made in answering the question.
First it is assumed that we can ignore air resistance.
Next, it is assumed that the given range is for the rock being launched and landing at the same height (i.e. on level ground).
Third, it is assumed that when he launches the rock to get maximum height (i.e. when he throws it vertically upward) that he can throw it with the same speed that he throws it to get maximum range (i.e. when he throws it at 45 degrees).
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.
Note: Some assumptions have to be made in answering the question.
First it is assumed that we can ignore air resistance.
Next, it is assumed that the given range is for the rock being launched and landing at the same height (i.e. on level ground).
Third, it is assumed that when he launches the rock to get maximum height (i.e. when he throws it vertically upward) that he can throw it with the same speed that he throws it to get maximum range (i.e. when he throws it at 45 degrees).
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