Question

A body is projected vertically up with ${\text{u}}$. What is its velocity at its maximum height?
(A) $\dfrac{u}{2}$
(B) $\dfrac{{{u^2}}}{2}$
(C) $\sqrt 2 u$
(D) $\dfrac{u}{{\sqrt 2 }}$

Answer 1

Use the formula for the maximum height attained by the projectile. Then use the kinematic equation relating final velocity, initial velocity, acceleration and displacement to substitute half of the maximum height for the displacement in the kinematic equation.

Formula used:
${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$
Here, u is the initial velocity of the projectile and g is the acceleration due to gravity.

Complete step by step answer:
We know that the maximum height attained by the projectile projected vertically upward is,
${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$ …… (1)
Here, u is the initial velocity of the projectile and g is the acceleration due to gravity.
Use the third kinematical equation to determine the velocity of the body at half of its maximum height as follows,
${v^2} = {u^2} + 2as$
Here, v is the final velocity of the body at its highest point, a is the acceleration of the body and s is the total upward distance covered by the body.
For a body projected in the upward direction, the acceleration produced in the body is acceleration due to gravity. Let upward direction be the positive direction, therefore, the acceleration due to gravity has the negative sign as it is acted in the downward direction.
Therefore, the above equation becomes,
${v^2} = {u^2} - 2gh$ …… (2)
Here, h is the maximum height attained by the body.
As we need to determine the velocity at half of the maximum height, substitute $h = \dfrac{{{H_{\max }}}}{2}$ in equation (1).
$h = \dfrac{{{u^2}}}{{4g}}$
Substitute $h = \dfrac{{{u^2}}}{{4g}}$ in equation (2).
${v^2} = {u^2} - 2g\dfrac{{{u^2}}}{{4g}}$
$\Rightarrow {v^2} = {u^2} - \dfrac{{{u^2}}}{2}$
$\Rightarrow {v^2} = \dfrac{{{u^2}}}{2}$
$\therefore v = \dfrac{u}{{\sqrt 2 }}$
So, the correct option is (D).

Note: In vertical projectile motion, the velocity of the body at its maximum height becomes zero. In this case, we are calculating the velocity of the body at half of the maximum height. Do not substitute 0 for v in the final equation.