Question

The velocity at the maximum height of a projectile is half of its initial velocity u. Its range on the horizontal plane is:
(A) $$$$$\dfrac{2{{u}^{2}}}{3g}$(B)$\dfrac{\sqrt{3}{{u}^{2}}}{2g}$(C)$\dfrac{{{u}^{2}}}{3g}$(D)$\dfrac{{{u}^{2}}}{2g}$

Answer 1

This is a case of projectile motion. Since the motion is two dimensional, the velocity has a horizontal and a vertical component. We have to find the maximum horizontal range it traverses, that is, its range. This is possible from calculation using the given conditions.

Formulas used:
Formula to find range of projectile motion:
$R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$ where R is the range of the body in projectile, u is the initial velocity, $\theta$ is the angle of projection and g is the acceleration due to gravity.

Complete step by step solution:
In our question, we are given that the velocity at the maximum height of a projectile is half of its initial velocity u. When a body is at its highest point during the projectile motion, the velocity acts only along the horizontal direction that is, only the horizontal component acts at that point of time. Hence, horizontal component ${{V}_{x}}=\dfrac{u}{2}$
Which implies that the vertical component ${{V}_{y}}=\sqrt{{{u}^{2}}-\dfrac{{{u}^{2}}}{4}}=\dfrac{u\sqrt{3}}{2}$
We know that
$\begin{aligned} & \tan \theta =\dfrac{{{V}_{y}}}{{{V}_{x}}}=\sqrt{3} \\\ & \Rightarrow \theta ={{60}^{\circ }} \\\ \end{aligned}$
From the equation to find the range of the projectile, that is
$R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$
When we substitute the values we obtain
$\therefore R=\dfrac{{{u}^{2}}\sin 2(60)}{g}=\dfrac{{{u}^{2}}\sin 120}{g}=\dfrac{\sqrt{3}{{u}^{2}}}{2g}$

Thus we can say that option B is the correct answer.

Note: In order to solve similar problems, students must remember the formulas for the time of flight, maximum height and the range of projectile as it comes handy while solving these problems. Also, the relations between these quantities are also to be remembered.