Question

The greatest height to which a man can through a stone is $h$. The greatest distance to which he can through it, will be:
A) $\dfrac{h}{2}$
B) $h$
C) $2h$
D) $3h$

Answer 1

Remember maximum height can be achieved by a thrown stone will be when the stone is thrown at angle of $90^\circ$ with the horizontal and the maximum distance covered by a thrown stone will be when the stone is thrown at angle of $45^\circ$ with the horizontal.

Complete step by step solution:
We know the maximum height can be achieved by a thrown stone if the stone is thrown at an angle of $90^\circ$ with the horizontal or vertically upwards.
It is given in the question that, the greatest height to which a man can through a stone is $h$
The height achieved by a thrown stone is given by, $h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where, $u$ is the velocity at which the stone is thrown,
$\theta$ is the angle with the horizontal at which the stone is thrown and
$g$ is the gravitational constant
The maximum height can be achieved will be at $90^\circ$ with the horizontal
That is, ${h_{\max }} = h = \dfrac{{{u^2}{{\sin }^2}90}}{{2g}}$
$\Rightarrow h = \dfrac{{{u^2} \times 1}}{{2g}}$
$\therefore {u^2} = h \times 2g$
We know the maximum distance covered by a thrown stone will be if the stone is thrown at angle of $45^\circ$ with the horizontal
That is $\theta = 45^\circ$
The range or the distance covered by a thrown stone, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$………………………. (1)
Where, $u$ is the velocity at which the stone is thrown
$\theta$ is the angle with the horizontal at which the stone is thrown
$g$ is the gravitational constant
Applying the value of ${u^2} = h \times 2g$ in equation (1) we get,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$\Rightarrow R = \dfrac{{2gh\sin (2 \times \theta )}}{g}$
$\Rightarrow R = \dfrac{{2gh\sin (2 \times 45)}}{g}$
$\Rightarrow R = \dfrac{{2gh\sin 90}}{g}$
$\therefore R = \dfrac{{2gh}}{g} = 2h$
That is the greatest distance to which the man can through it, will be: $R = 2h$

So the final answer is Option (C), $2h$.

Note: The stone’s maximum vertical displacement will occur when it is thrown vertically upwards with some initial velocity. The stone will have an initial kinetic energy and this kinetic energy will be zero at the topmost point of the motion. Thus, the velocity of the stone at the topmost point is zero. From that point the stone will undergo free fall motion. Thus regaining the lost kinetic energy. The initial velocity of the stone in the case of the projectile motion is the same as that of the vertical throw. Hence we get the value of maximum horizontal distance in terms of the maximum vertical distance.