Question

A projectile is thrown from a point in a vertical plane such that its horizontal and vertical velocity component are $9.8\,m{s^{ - 1}}$ and $19.6\,m{s^{ - 1}}$ respectively. Its horizontal range is :
A. $4.9\,m$
B. $9.8\,m$
C. $19.6\,m$
D. $39.2\,m$

Answer 1

In this problem, we have to use the basic formulae and find out the initial velocity and the angle of projection. Then we can proceed to put it in the formula for horizontal range of a projectile and get the required answer.

Formulae used:
$v = \sqrt {{v_x}^2 + {v_y}^2}$
where $v$ is the velocity, ${v_x}$ is the horizontal component of the velocity and ${v_y}$ is the vertical component of the velocity.
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$
where $R$ is the Range, $v$ is the initial velocity, $g$ is the acceleration due to gravity and $\theta$ is the angle of projection of the projectile.

Complete step by step answer:
According to the given question
Horizontal component of velocity ${v_x} = 9.6\,m{s^{ - 1}}$
Vertical component of velocity ${v_y} = 19.6\,m{s^{ - 1}}$
Now, to find out the initial velocity, we put the values of ${v_x}$ and ${v_y}$ in the equation
$v = \sqrt {{v_x}^2 + {v_y}^2}$
$\Rightarrow v = \sqrt {{{(9.8)}^2} + {{(19.6)}^2}} \\\ \Rightarrow v = 9.8\sqrt 5 \\\$
We know that
${v_x} = v\sin \theta$
$\Rightarrow {v_y} = v\cos \theta$
Substituting the values of ${v_x}$ , ${v_y}$ and $v$ in the above equations, we get
$\Rightarrow 9.8 = 9.8\sqrt 5 \sin \theta \\\ \Rightarrow \sin \theta = \dfrac{1}{{\sqrt 5 }}$
And
$\Rightarrow 19.6 = 9.8\sqrt 5 \cos \theta \\\ \Rightarrow \cos \theta = \dfrac{2}{{\sqrt 5 }}$
Now, using the formula for finding range, we get
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$

Now, using the trigonometric transformation $\sin 2\theta = 2\sin \theta \cos \theta$ , we get
$R = \dfrac{{{v^2}2\sin \theta \cos \theta }}{g}$
Thereafter putting in the respective values in this formula, we get
$\Rightarrow R = \dfrac{{{{(9.8\sqrt 5 )}^2}2\left( {\dfrac{1}{{\sqrt 5 }}} \right)\left( {\dfrac{2}{{\sqrt 5 }}} \right)}}{{9.8}} \\\ \Rightarrow R = 9.8 \times 4 \\\ \therefore R = 39.2m \\\$
Thus, the correct answer is option D.

Note: Acceleration in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.