Question

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

A

$\sqrt{2u}$

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B

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

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C

$\frac{u}{2d}$

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D

$\frac{{u}^{2}}{2}$

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Solution

## The correct option is B $\frac{u}{\sqrt{2}}$Step 1: Given dataA body is projected vertically up with velocity uStep 2: Formula used${H}_{max}=\frac{{u}^{2}}{2g}\left[where{H}_{max}=maximumheightattainedbytheprojectile,u=initialvelocity,g=accelerationduetogravity\right]$Step 3: Calculating velocity at half its maximum heightUse the third kinematical equation to determine the velocity of the body at half of its maximum height as follows,${v}^{2}={u}^{2}+2aswherev=finalvelocityatitshighestpoint,u=initialvelocity,a=acceleration,s=totalupwarddis\mathrm{tan}ce\right]$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 ${v}^{2}={u}^{2}-2gh\left[h=maximumheightattained\right]............\left(i\right)$As we need to determine the velocity at half of the maximum height, substitute $h=\frac{{H}_{max}}{2}$ ${H}_{max}=\frac{{u}^{2}}{2g}\phantom{\rule{0ex}{0ex}}h=\frac{{u}^{2}}{4g}$Substituting the value of $h=\frac{{u}^{2}}{4g}$ in equation (i)${v}^{2}={u}^{2}-2g\frac{{u}^{2}}{4g}\phantom{\rule{0ex}{0ex}}{v}^{2}={u}^{2}-\frac{{u}^{2}}{2}\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{{u}^{2}}{2}\phantom{\rule{0ex}{0ex}}v=\frac{u}{\sqrt{2}}$Hence, option B is the correct answer.

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