Question

A projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at height equal to half of the maximum height, then the vertical component of the velocity of projectile is:
A. $3v\sin \theta$
B. $\dfrac{{v\sin \theta }}{3}$
C. $\dfrac{{v\sin \theta }}{{\sqrt 2 }}$
D. $\dfrac{{v\sin \theta }}{{\sqrt 3 }}$

Answer 1

For solving this problem, we need to use the equation of maximum height of the projectile. Maximum height is the trajectory’s highest point where the vertical component of the velocity is zero. But here, we are asked to find the vertical component of the velocity when the projectile is at height equal to half of the maximum height. Therefore, we will also use one of the differential equations of motion.

Formulas used:
${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where, ${H_{\max }}$ is the maximum height of the projectile, $u$ is the initial velocity by which the projectile is thrown, $\theta$ is an angle with which the projectile is thrown with horizontal and $g$ is the gravitational acceleration.
$s = ut + \dfrac{1}{2}g{t^2}$
Where, $s$ is the distance covered by the projectile, $u$ is the initial velocity by which the projectile is thrown, $t$ is the time taken by the projectile to cover the distance $s$ and $g$ is the gravitational acceleration.

Complete step by step solution:
We know that the maximum height covered by the projectile is given by:
${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Here we are given that the initial velocity is $v$ by which the projectile is thrown
Therefore, ${H_{\max }} = \dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}$
Also, the horizontal component of velocity ${v_x} = v\cos \theta$ and vertical component ${v_y} = v\sin \theta$.
Now let us find the time taken by the projectile to reach at half of the maximum height by using the equation,
$s = ut + \dfrac{1}{2}g{t^2}$
Here, we will take $s = \dfrac{{{H_{\max }}}}{2}$ , $u = {v_y}$ and gravitational acceleration as $- g$ as the projectile is going in upward direction initially.

$$\dfrac{{{H_{\max }}}}{2} = {v_y}t - \dfrac{1}{2}g{t^2} \\\ \Rightarrow \dfrac{{{v^2}{{\sin }^2}\theta }}{{4g}} = v\sin \theta \cdot t - \dfrac{1}{2}g{t^2} \\\ \Rightarrow g = 10m/{s^2} \\\ \Rightarrow \dfrac{{{v^2}{{\sin }^2}\theta }}{{40}} = v\sin \theta \cdot t - 5{t^2} \\\ \Rightarrow 200{t^2} - 40v\sin \theta \cdot t + {v^2}{\sin ^2}\theta = 0 \\\ \Rightarrow t = \dfrac{{\left( {\sqrt 2 \pm 1} \right)v\sin \theta }}{{10\sqrt 2 }} \\\$$

Now, the vertical component of velocity at half of the maximum height can be found by using the equation,
${v_y}^1 = v\sin \theta - \dfrac{{\left( {\sqrt 2 - 1} \right)v\sin \theta }}{{\sqrt 2 }} \\\ \Rightarrow {v_y}^1 = \dfrac{{v\sin \theta }}{{\sqrt 2 }}$
$\Rightarrow{v_y}^1 = {v_y} - gt$
Putting the value of $t$, we get
${v_y}^1 = v\sin \theta - \dfrac{{\left( {\sqrt 2 - 1} \right)v\sin \theta }}{{\sqrt 2 }} \\\ \therefore {v_y}^1 = \dfrac{{v\sin \theta }}{{\sqrt 2 }}$
Hence, option C is the correct answer.

Note: Here, we have determined the vertical component of the velocity at the height half of the maximum height. Similarly at any given height up to the maximum height, we can find the vertical component of the velocity by following the same procedure.