Question

A particle is projected from the ground with an initial speed of $v$ at an angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is?
A) $\dfrac{v}{2}\sqrt {1 + 2{{\cos }^2}\theta }$
B) $\dfrac{v}{2}\sqrt {1 + {{\cos }^2}\theta }$
C) $\dfrac{v}{2}\sqrt {1 + 3{{\cos }^2}\theta }$
D) $v\cos \theta$

Answer 1

In order to find the correct solution of the given question first of all we need to know the formula for the height of the highest point of the projectile. After that we need to find the total horizontal distance covered and then find the total displacement of the projectile. Then we can easily find the value of the required average velocity and finally conclude with the correct solution of the given question.

Complete step by step solution:
We know that the height of a projectile at its highest point is given as $h = \dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}$
Also, we know that the horizontal range of the projectile is given as $R = \dfrac{{{v^2}\sin 2\theta }}{g}$
Now, the horizontal distance covered till the highest point is given by $x = \dfrac{R}{2} = \dfrac{{{v^2}\sin 2\theta }}{{2g}} = \dfrac{{{v^2}\sin \theta \cos \theta }}{g}$
Now, we need to find the time taken to reach the highest point. So, we can find it as,
$\Rightarrow t = \dfrac{T}{2} = \dfrac{{2v\sin \theta }}{{2g}} = \dfrac{{v\sin \theta }}{g}$
Now, in order to find the average velocity, we need to find the displacement of the projectile at the highest point which can be written as,
$\Rightarrow d = \sqrt {{h^2} + {x^2}} = \sqrt {{{\left( {\dfrac{{{v^2}\sin 2\theta }}{{2g}}} \right)}^2} + {{\left( {\dfrac{{{v^2}\sin \theta \cos \theta }}{g}} \right)}^2}}$
$\Rightarrow d = \dfrac{{{v^2}\sin \theta }}{{2g}}\sqrt {1 + 3{{\cos }^2}\theta }$
Now, we need to find the average velocity. So, it can be written as,
$\Rightarrow {v_{av}} = \dfrac{d}{t}$
After putting the values of the displacement and time in the above equation, we get,
$\Rightarrow {v_{av}} = \dfrac{{\dfrac{{{v^2}\sin \theta }}{{2g}}\sqrt {1 + 3{{\cos }^2}\theta } }}{{\dfrac{{v\sin \theta }}{g}}}$
$\therefore {v_{av}} = \dfrac{v}{2}\sqrt {1 + 3{{\cos }^2}\theta }$

Hence, option (C) is the correct choice for the given question.

Note: We define the range of the projectile as the horizontal distance covered by the body when it is in motion. The body can cover maximum distance if it is projected at an angle of${45^\circ }$. The time taken by a body when it is projected and lands is called the time of flight. We should also know this fact that the time of flight depends on the initial velocity and the angle of projection of the projectile.