Question

The range of a projectile when fired at $ {75^ \circ } $ with the horizontal is $ 0.5km $ . What will be its range when fired at $ {45^ \circ } $ with the same speed?
 $ \left( A \right){\text{ 0}}{\text{.5km}} $
 $ \left( B \right){\text{ 1}}{\text{.0km}} $
 $ \left( C \right){\text{ 1}}{\text{.5km}} $
 $ \left( D \right){\text{ 20km}} $

Answer 1

Hint: In this question, we have to find the range of the projectile at a certain angle, and as we know that the formula of the range is given by $ \dfrac{{{u^2}\sin 2\theta }}{g} $ . So by substituting the angle and solving for it we will get the value for the range of the projectile.

Formula used
Range of projectile is given by,
 $ R = \dfrac{{{u^2}\sin 2\theta }}{g} $
Here, $ R $ , will be the range,
 $ u $ , will be the initial velocity,
 $ g $ , will be the acceleration due to gravity.

Complete Step By Step Answer:
So the horizontal range is given by
 $ \Rightarrow \dfrac{{{u^2}\sin 2\theta }}{g} = 0.5km $
Since we have an angle $ {75^ \circ } $ , so on substituting the values, we will get the equation as
 $ \Rightarrow \dfrac{{{u^2}\sin \left( {2 \times {{75}^ \circ }} \right)}}{g} = 0.5km $
And on solving the above equation we will get the equation as
 $ \Rightarrow \dfrac{{{u^2}\sin \left( {{{150}^ \circ }} \right)}}{g} = 0.5km $
And for solving it, the degree can be written as
 $ \Rightarrow \dfrac{{{u^2}\sin \left( {{{180}^ \circ } - {{30}^ \circ }} \right)}}{g} = 0.5km $
Now on solving it, we get
 $ \Rightarrow \dfrac{{{u^2}}}{{2g}} = 0.5km $
Therefore, the range $ {R_1} = \Rightarrow \dfrac{{{u^2}}}{{2g}} = 0.5km $
Now for the range $ {R_2} $ , we have an angle $ {45^ \circ } $ , we get
On substituting the values, we will get the equation as
 $ \Rightarrow \dfrac{{{u^2}\sin \left( {2 \times {{45}^ \circ }} \right)}}{g} $
And on solving the multiplication, we get
 $ \Rightarrow \dfrac{{{u^2}\sin \left( {{{90}^ \circ }} \right)}}{g} $
And by solving it we get
 $ \Rightarrow \dfrac{{{u^2}}}{g} $
And as we know that the above value is just the double of $ {R_1} $
Hence, $ {R_2} = 2{R_1} $
And on substituting the values, we get
 $ \Rightarrow {R_2} = 2 \times 0.5 $
And on solving it, we get
 $ \Rightarrow {R_2} = 1km $
Hence, the range when fired at $ {45^ \circ } $ with the same speed will be equal to $ 1km $ .

Note:
We should know that the initial velocity and angle of projection will choose the range as well as the height it will attain. Also while solving the trigonometric angles we should first make the angle in such a way that it follows some identity and then we can easily solve it