Question

In an oblique projectile motion, if the velocity of projection is increased by $2\% $, the percentage increase in horizontal range will be
(A) $1\% $
(B) $2\% $
(C) $3\% $
(D) $4\% $

Answer 1

Hint: When an object is projected into the air with a velocity, it is called a projectile. The projectile moves under the influence of the gravity of the earth. The path of a projectile will be a parabola. We have to find the percentage increase in the horizontal range of the projectile when the velocity of the projectile is increased by $2\% $.
Formula used:
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ $\left( {\because \sin 2\theta = 2\sin \theta \cos \theta } \right)$
Where $R$ stands for the horizontal range of the projectile, $$$u\cos \theta $stands for the horizontal velocity of the projectile, $\dfrac{{2u\sin \theta }}{g}$stands for the time of flight of the projectile.

Complete step by step answer:
Horizontal range of a projectile is the distance between the point of projection and the point where the trajectory meets the horizontal line through the point of projection.
The horizontal range of a projectile is given by,
$\Rightarrow$ $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
When the velocity of the projectile is increased by $2\% $, the horizontal range will become
\[R' = \dfrac{{{u^2} + {{\left( {\dfrac{{2u}}{{100}}} \right)}^2}}}{g}\sin 2\theta \]
This will become,
$\Rightarrow$ $R' = \dfrac{{{u^2}{{\left( {1 + \dfrac{2}{{100}}} \right)}^2}}}{g}\sin 2\theta $
Taking the ratio of $R'$ and $R$, we get
$\Rightarrow$ $\dfrac{{R'}}{R} = \dfrac{{\dfrac{{{u^2}{{\left( {1 + \dfrac{2}{{100}}} \right)}^2}}}{g}\sin 2\theta }}{{\dfrac{{{u^2}\sin 2\theta }}{g}}}$
Canceling common terms, we get
$\Rightarrow$ $\dfrac{{R'}}{R} = {\left( {1 + \dfrac{2}{{100}}} \right)^2}$
On simplifying we get,
$\Rightarrow$ $\dfrac{{R'}}{R} = {1^2} + {\left( {\dfrac{2}{{100}}} \right)^2} + \left( {2 \times \dfrac{2}{{100}}} \right)$
We know that, ${\left( {\dfrac{2}{{100}}} \right)^2} \ll 1$, we can neglect the term.
Therefore,
$\Rightarrow$ $\dfrac{{R'}}{R} = 1 + \dfrac{4}{{100}}$
Subtracting $1$ from both sides,
$\Rightarrow$ $\dfrac{{R'}}{R} - 1 = \dfrac{4}{{100}}$
This can be written as,
$\Rightarrow$ $\dfrac{{R' - R}}{R} = \dfrac{4}{{100}}$
This can be written as,
$\Rightarrow$ $\dfrac{{\Delta R}}{R} = \dfrac{4}{{100}}$
The percentage increase in horizontal range will be
$\Rightarrow$ $\dfrac{{\Delta R}}{R} \times 100 = 4$

The answer is: Option (D): $4\% $

Note:
Alternative method:
We know that the horizontal range is directly proportional to the square of the velocity of the projectile,
i.e.
$\Rightarrow$ $R \propto {u^2}$
This can be written as,
$\Rightarrow$ $R = k{u^2}$
The percentage change in velocity can be written as,
$\Rightarrow$ $\dfrac{{\Delta u}}{u} \times 100 = 2\% $
The horizontal range can be written as,
$\Rightarrow$ $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
The percentage change in the horizontal range can be written as,
$\Rightarrow$ $\dfrac{{\Delta R}}{R} \times 100 = 2 \times \dfrac{{\Delta u}}{u} \times 100$
Substituting, we get
$\Rightarrow$ $\dfrac{{\Delta R}}{R} \times 100 = 2 \times 2\% = 4\% $