A body is projected vertically upwards with a velocity of 10,ms^{-1}. It reaches maximum vertical he

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2.Motion in Straight Line

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A body is projected vertically upwards with a velocity of $10\,ms^{-1}.$ It reaches the maximum vertical height $h$ in time $t.$ In $\frac {t}{2}$, the height covered is

A

$\frac {h}{2}$

B

$\frac {2h}{5}$

C

$\frac {3h}{4}$

D

$\frac {5h}{8}$

Solution

$\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2} \quad \ldots(\mathrm{i})$

and $v=g t . .(\text { ii })$

$h^{\prime}=v \frac{t}{2}-\frac{1}{2} g\left(\frac{t}{2}\right)^{2}=g t \frac{t}{2}-\frac{1}{4} \times \frac{1}{2} g t^{2}$

$=h-\frac{1}{4} h=\frac{3}{4} h$

Standard 11

Physics

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