Question

Two stones are projected from the same point with the same speed making angles $45^\circ + x$ and $45^\circ - x$ with the horizontal respectively. If $x \leqslant 45^\circ $ then the horizontal ranges of the two stones are in the ratio of?

Answer 1

Hint: In this we have two stones that are projected at the same speed making a different angle with the horizontal. Now using the projectile motion horizontal range formula for two stones we will get the relation between the horizontal range, horizontal angle, projected speed of the stone and the acceleration due to gravity. After that taking the ratio of the horizontal ranges we will get our solution.

Complete step by step answer:
As per the problem we have two stones which are projected from the same point with the same speed making angles $45^\circ + x$ and $45^\circ - x$ with the horizontal respectively and $x \leqslant 45^\circ $. We have to find the ratio of horizontal ranges of the two stones.We know,
${u_1} = {u_2}$
Where ${u_1}\,\,and\,\,{u_2}$ are the speed of the stone one and two respectively.
We know the formula of horizontal range of a projectile motion,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Now the horizontal range for the stone one is,
${R_1} = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}$
We know, ${\theta _1} = 45^\circ + x$

Now putting the ${\theta _1}$ value in the range we will get,
${R_1} = \dfrac{{{u^2}\sin 2\left( {45^\circ + x} \right)}}{g}$
$ \Rightarrow {R_1} = \dfrac{{{u^2}\sin \left( {90^\circ + 2x} \right)}}{g}$
We know that, $\sin \left( {90^\circ + 2\theta } \right) = \cos 2\theta $
Now,
${R_1} = \dfrac{{{u^2}\cos 2x}}{g} \ldots \ldots \left( 1 \right)$
Similarly for stone two
${R_2} = \dfrac{{{u^2}\sin 2{\theta _2}}}{g}$
We know,
${\theta _2} = 45^\circ - x$

Now putting the ${\theta _2}$ value in the range we will get,
${R_2} = \dfrac{{{u^2}\sin 2\left( {45^\circ - x} \right)}}{g}$
$ \Rightarrow {R_2} = \dfrac{{{u^2}\sin \left( {90^\circ - 2x} \right)}}{g}$
We know that, $\sin \left( {90^\circ - 2\theta } \right) = \cos 2\theta $
Now,
${R_2} = \dfrac{{{u^2}\cos 2x}}{g} \ldots \ldots \left( 2 \right)$
Now taking ratio of equation $\left( 1 \right)$ to $\left( 2 \right)$ we will get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{{u^2}\cos 2x}}{g}}}{{\dfrac{{{u^2}\cos 2x}}{g}}}$
Cancelling all the common terms we will get,
$\dfrac{{{R_1}}}{{{R_2}}} = 1$

Hence, ${R_1}:{R_2} = 1:1$.

Note: Here the horizontal range is defined as the distance along the horizontal plane or we can say along the x direction by the particle. Moreover, the particle would travel before reaching the same vertical position as it started from. And the horizontal range of the particle depends on the launch angle, launch speed and acceleration of gravity.