Question

If a person can through a stone upto a maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is,
A) $\dfrac{h}{2}$metres.
B) h metres.
C) 2h metres.
D) 3h metres.

Answer 1

Hint
Find out the relation between u and H using the condition of Maximum vertical height. At maximum height $v = 0$. And then for maximum horizontal range use the formula for range which is given by $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ and put the value of $\theta \,{\text{as}}\,{45^{^ \circ }}$and solve for R to find out the answer.

Complete step by step answer
Let the person throw the stone with an initial velocity u and when the stone reaches the maximum height, its final velocity becomes zero. So, using the third law of motion we have,
${v^2} = {u^2} + 2aS$
Putting the values as mentioned we have,
$\Rightarrow {0^2} = {u^2} - 2gH$
$\Rightarrow H = \dfrac{{{u^2}}}{{2g}}$
$\Rightarrow {u^2} = 2gH$
Now for the maximum distance to be covered (horizontal) the angle with which the stone should be thrown must be ${45^{^ \circ }}$.
In case of a projectile to attain maximum range, angle must be ${45^{^ \circ }}$.
So, the maximum horizontal range is given by,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
On putting the appropriate values we have,
$\Rightarrow R = \dfrac{{{u^2}\sin (2 \times {{45}^{^ \circ }})}}{g}$
$\Rightarrow R = \dfrac{{{u^2}\sin ({{90}^{^ \circ }})}}{g}$
On further solving we have,
$\Rightarrow R = \dfrac{{2gH}}{g}$
$\Rightarrow R = 2H$
Therefore the maximum range through which the stone travels is 2H.
Hence the correct answer is option (C).

Note
At both angles $\theta \,{\text{and}}\,(90 - \theta )$the horizontal range is the same. This can be proved by inverting the value of $\theta$as $\theta \,{\text{and}}\,(90 - \theta )$in the formula, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$. By using the simple trigonometric identity like $\sin 2(90 - \theta ) = \sin (180 - 2\theta ) = \sin 2\theta$, we can prove the following statement.