A body thrown vertically up with initial velocity 52 ,m / s from ground passes twice a point at h he

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2.Motion in Straight Line

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A body thrown vertically up with initial velocity $52 \,m / s$ from the ground passes twice a point at $h$ height above at an interval of $10 \,s$. The height $h$ is .........$m$ $\left(g=10 \,m / s ^2\right)$

A

$22$

B

$10.2$

C

$11.2$

D

$15$

Solution

(b)

Given, $t_2-t_1=10 s$

$t_2+t_1=\frac{2 u}{g}=\frac{2 \times 52}{10}=10.4$

$\Rightarrow 2 t_2=20.4$

$\Rightarrow t_2=10.2 \,s$

$t_1=0.2 \,s$

So, $t_1 t_2=\frac{2 h}{g}$

$0.2 \times 10.2=\frac{2 \times h}{10}$

$\Rightarrow 1 \times 10.2=h \Rightarrow 10.2 \,m =h$

Standard 11

Physics

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