Question

A projectile is fired making an angle $2\theta$ with the horizontal with a velocity of $4m/s$ . At a particular instant, it makes an angle $\theta$ with the horizontal. What is its velocity at that instant?
(A) $4\cos \theta$
(B) $4\left( {\sec \theta + \cos \theta } \right)$
(C) $2\left( {\sec \theta + 4\cos \theta } \right)$
(D) $4\left( {2\cos \theta - \sec \theta } \right)$

Answer 1

We can decide the initial velocity and velocity at a definite point on trajectory into their horizontal and vertical components. Now by using basic trigonometric relations between the components, we can determine the required answer.

Complete step by step answer:
When an object is thrown with some initial velocity and is permitted to fall freely under the effect of gravity of earth, then the object follows a curved path. The object is well-known as a projectile so this type of motion exhibited by an object under gravity is called the projectile motion.
When a projectile is projected with some initial velocity $u$ now as it moves along its path, its velocity keeps changing and is not fixed at $u$ .
In this question, we have a particle which is projected with a velocity $u$ and makes an angle $2\theta$ with the horizontal.
We can decide the initial velocity into its horizontal and vertical components.
We are given that at every point of the trajectory, the velocity $v$ of the particle perpendicular to its initial velocity by reproducing the directions of $u$ and $v$ meeting at $\theta$ .
Now we see that the horizontal components of $u$ and $v$ must be equal. And so, we can write the following expression for the particle.
$v\cos \theta = u\cos 2\theta$
$\Rightarrow v = \dfrac{{u\cos 2\theta }}{{\cos \theta }}$
Where, $\sec \theta = \dfrac{1}{{\cos \theta }}$
$\therefore v = u\cos 2\theta \sec \theta$
Using the trigonometric identity,
$\cos 2\theta = 2{\cos ^2}\theta - 1$
We have given the initial velocity is $u = 4m/s$ .
We get $v = 4\left( {2{{\cos }^2}\theta - 1} \right)\sec \theta$
On further solving the above equation we get,
$v = 4\left( {2\cos \theta - \sec \theta } \right)$
Therefore, the velocity at a particular instant when the projectile makes an angle $\theta$ with the horizontal is $4\left( {2\cos \theta - \sec \theta } \right)$ .
Hence, the correct answer is option (D).

Note:
The velocity of the projectile retains changing along its trajectory. As the projectile rises above the ground, its velocity keeps on decreasing. At the highest point $v = 0$ . As the projectile starts falling, $v$ starts increasing but the final velocity is zero as the projectile comes to rest when it completes its trajectory.