Question

A body is projected vertically up with velocity u. Its velocity at half its maximum height is?

• A. $\sqrt{2u}$
• B. $\frac{u}{\sqrt{2}}$
• C. $\frac{u}{2d}$
• D. $\frac{u^2}{2}$

Answer 1

The correct option is B

$\frac{u}{\sqrt{2}}$

Step 1: Given data

A body is projected vertically up with velocity u

Step 2: Formula used

$H_{max}=\frac{u^2}{2g}[where{H}_{max}=maximumheightattainedbytheprojectile,u=initialvelocity,g=accelerationduetogravity]$

Step 3: Calculating velocity at half its maximum height

Use the third kinematical equation to determine the velocity of the body at half of its maximum height as follows,

$v^2=u^2+2aswherev=finalvelocityatitshighestpoint,u=initialvelocity,a=acceleration,s=totalupwarddis\tan ce]$

The acceleration produced in a body propelled upward is called acceleration owing to gravity. Assuming that the upward direction is positive, the acceleration owing to gravity has a negative sign because it acts in the downward direction.

Therefore, the above equation becomes $v^2=u^2-2gh[h=maximumheightattained]............\left(i\right)$

As we need to determine the velocity at half of the maximum height, substitute $h=\frac{H_{max}}{2}$

$$\begin{gathered} H_{max}=\frac{u^2}{2g} \\ h=\frac{u^2}{4g} \end{gathered}$$

Substituting the value of $h=\frac{u^2}{4g}$ in equation (i)

$$\begin{gathered} v^2=u^2-2g\frac{u^2}{4g} \\ {v}^2=u^2-\frac{u^2}{2} \\ {v}^2=\frac{u^2}{2} \\ v=\frac{u}{\sqrt{2}} \end{gathered}$$

Hence, option B is the correct answer.