Question

For elevations which exceed or fall short of ${45^ \circ }$ by equal amounts the ranges are equal?
(A) True
(B) False

Answer 1

Hint:For the elevation kinds of the problem and the angle is given in the question, the solution can be determined by using the projectile motion formula, assume one angle and that angle is added to the given angle in the question and then calculating the range value and the assumed angle is subtracted to the given angle in the question and then the range value is calculated, then by comparing the two range values, the solution is determined.

Useful formula:
From the projectile motion equation, the range is given as,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where, $R$ is the range, $u$ is the velocity, $\theta $ is the angle and $g$ is the acceleration due to gravity.

Complete step by step solution:
Given that,
For elevations which exceed or fall short of ${45^ \circ }$.
Assume the angle as $\alpha $
Then, ${\theta _1} = {45^ \circ } + \alpha $
${\theta _2} = {45^ \circ } - \alpha $
 Now,
The range is given as,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}\,...................\left( 1 \right)$
For the angle ${\theta _1}$, then the range is given by,
${R_1} = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}\,...............\left( 2 \right)$
By substituting the angle ${\theta _1}$ in the above equation (2) then,
${R_1} = \dfrac{{{u^2}\sin 2\left( {{{45}^ \circ } + \alpha } \right)}}{g}$
On multiplying the above equation, then the above equation is written as,
${R_1} = \dfrac{{{u^2}\sin \left( {{{90}^ \circ } + 2\alpha } \right)}}{g}$
From trigonometry the $\sin \left( {{{90}^ \circ } + \alpha } \right) = \cos \alpha $, by substituting this in the above equation, then
${R_1} = \dfrac{{{u^2}\cos 2\alpha }}{g}\,....................\left( 3 \right)$
For the angle ${\theta _2}$, then the range is given by,
${R_2} = \dfrac{{{u^2}\sin 2{\theta _2}}}{g}\,................\left( 4 \right)$
By substituting the angle ${\theta _2}$ in the above equation (4) then,
${R_2} = \dfrac{{{u^2}\sin 2\left( {{{45}^ \circ } - \alpha } \right)}}{g}$
On multiplying the above equation, then the above equation is written as,
${R_2} = \dfrac{{{u^2}\sin \left( {{{90}^ \circ } - 2\alpha } \right)}}{g}$
From trigonometry the $\sin \left( {{{90}^ \circ } - \alpha } \right) = \cos \alpha $, by substituting this in the above equation, then
${R_2} = \dfrac{{{u^2}\cos 2\alpha }}{g}\,.....................\left( 5 \right)$
From the equation (3) and equation (5),
${R_1} = {R_2}$

Hence, the option (A) is the correct answer.

Note: For solving this question some trigonometric equation must be known like $\sin \left( {{{90}^ \circ } + \alpha } \right) = \cos \alpha $. And in equation (3), the angle is written as $\cos 2\alpha $, because in the above step the value $2$ is multiplied with the angle $\alpha $, so the angle is written as $\cos 2\alpha $.