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Question: A person can throw a stone to a maximum distance of 100 m. The greatest height to which he can throw...

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

Explanation

Solution

The maximum distance given, 100m, is the range of projectile motion. The maximum height is achieved when thrown straight up at 45{45^ \circ } angle. Thus launching angle is θ=45\theta = {45^ \circ }.
Formula Used: The formulae used in the solution are given here.
Range is given as R=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g} where uu is the initial velocity, θ\theta is the launch angle and gg is the acceleration due to gravity and the maximum height is given by H=u2sin2θ2gH = \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=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g} where uu is the initial velocity, θ\theta is the launch angle and gg is the acceleration due to gravity.
Since, R=100mR = 100m, thus u2sin2θg=100\dfrac{{{u^2}\sin 2\theta }}{g} = 100 and the acceleration due to gravity g=9.8g = 9.8.
Thus, it can be written that, u2sin2θ=100g{u^2}\sin 2\theta = 100g where g=9.8g = 9.8
It is known to us that, the greatest range is achieved using a 45{45^ \circ } launch angle:
u2sin(2×45)=980{u^2}\sin \left( {2 \times {{45}^ \circ }} \right) = 980
u2sin90=980\Rightarrow {u^2}\sin {90^ \circ } = 980
We know that, sin90=1\sin {90^ \circ } = 1,
u2=980\therefore {u^2} = 980
The height is achieved when thrown straight up at 45{45^ \circ } angle.
Thus, height H=u2sin2θ2gH = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}.
Since, u2=980{u^2} = 980, θ=45\theta = {45^ \circ } and g=9.8g = 9.8
H=98039.2H = \dfrac{{980}}{{39.2}}
H=25m\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).