Question

At what angle (in degrees) with the horizontal should a ball be thrown so that its range R is related to the time of flight as ${\text{R}} = {\text{5}}{{\text{T}}^2}$ (Take ${\text{g}} = 10{\text{m}}{{\text{s}}^{ - 2}}$)?
${\text{A}}{\text{.}}$ 30
${\text{B}}{\text{.}}$ 45
${\text{C}}{\text{.}}$ 60
${\text{D}}{\text{.}}$ 90

Answer 1

Hint: Here, we will proceed by putting the expressions for the range and the time of flight of an object undergoing projectile motion in the given equation. Then, we will be simplifying that equation.
Formulas Used: ${\text{R}} = \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}}$, ${\text{T}} = \dfrac{{{\text{2u}}\sin \theta }}{{\text{g}}}$, $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ and $\tan {45^0} = 1$.

Complete Step-by-Step solution:
Let the required angle which the ball will be making with the horizontal be $\theta $
Let u be the initial velocity with which the ball is thrown,
Given, ${\text{R}} = {\text{5}}{{\text{T}}^2}{\text{ }} \to {\text{(1)}}$ where R denotes the range of the projectile motion of thrown ball and T denotes the time of flight of this projectile motion
As we know that for any projectile motion,
Range, ${\text{R}} = \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}}{\text{ }} \to {\text{(2)}}$ where u denotes the initial velocity by which the object undergoing projectile motion is thrown and $\theta $ is the angle made (with the horizontal) by that object.
 Time of flight, ${\text{T}} = \dfrac{{{\text{2u}}\sin \theta }}{{\text{g}}}{\text{ }} \to {\text{(3)}}$ where u denotes the initial velocity by which the object undergoing projectile motion is thrown and $\theta $ is the angle made (with the horizontal) by that object.
Here, the object undergoing projectile is the ball.
By substituting equations (2) and (3) in equation (1), we get
$
   \Rightarrow \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}} = {\text{5}}{\left[ {\dfrac{{{\text{2u}}\sin \theta }}{{\text{g}}}} \right]^2} \\
   \Rightarrow \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}} = {\text{5}}\left[ {\dfrac{{{\text{4}}{{\text{u}}^2}{{\left( {\sin \theta } \right)}^2}}}{{{{\text{g}}^2}}}} \right] \\
   \Rightarrow \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}} = \dfrac{{{\text{20}}{{\text{u}}^2}{{\left( {\sin \theta } \right)}^2}}}{{{{\text{g}}^2}}} \\
 $
Using the formula $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ in the above equation, we get
\[
   \Rightarrow \dfrac{{{{\text{u}}^2}\left[ {2\left( {\sin \theta } \right)\left( {\cos \theta } \right)} \right]}}{{\text{g}}} = \dfrac{{{\text{20}}{{\text{u}}^2}{{\left( {\sin \theta } \right)}^2}}}{{{{\text{g}}^2}}} \\
   \Rightarrow \dfrac{{2{{\text{u}}^2}\left( {\sin \theta } \right)\left( {\cos \theta } \right)}}{{\text{g}}} = \dfrac{{{\text{20}}{{\text{u}}^2}{{\left( {\sin \theta } \right)}^2}}}{{{{\text{g}}^2}}} \\
 \]
By cancelling $\dfrac{{{\text{2}}{{\text{u}}^2}\left( {\sin \theta } \right)}}{{\text{g}}}$ from both sides of the above equation, we get
$ \Rightarrow \dfrac{{\left( {\cos \theta } \right)}}{{\text{1}}} = \dfrac{{{\text{10}}\left( {\sin \theta } \right)}}{{\text{g}}}$
By applying cross multiplication in the above equation, we get
$
   \Rightarrow {\text{g}}\left( {\cos \theta } \right) = 1{\text{0}}\left( {\sin \theta } \right) \\
   \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\text{g}}}{{10}} \\
 $
Using the formula $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ in the above equation, we get
$ \Rightarrow \tan \theta = \dfrac{{\text{g}}}{{10}}$
By substituting ${\text{g}} = 10{\text{m}}{{\text{s}}^{ - 2}}$ in the above equation, we get
$
   \Rightarrow \tan \theta = \dfrac{{{\text{10}}}}{{10}} \\
   \Rightarrow \tan \theta = 1{\text{ }} \to {\text{(4)}} \\
 $
According to the general trigonometric table, we can write
$\tan {45^0} = 1{\text{ }} \to {\text{(5)}}$
By comparing equations (4) and (5), we have
$\theta = {45^0}$
Therefore, the required angle of the ball made with the horizontal such that the given relation is satisfied is ${45^0}$.
Hence, option B is correct.

Note- Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth.