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

Hint: 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.