A ball is thrown vertically upwards from ground G with a speed u . It reaches a point B at a height

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2.Motion in Straight Line

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A ball is thrown vertically upwards from the ground $G$ with a speed $u$. It reaches a point $B$ at a height $h$ (lower than the maximum height) after time $t_1$. It returns to the ground after time $t_2$ from the instant it was at $B$ during the upward journey. Then $t_1t_2$ is equal to

A

$2h/g$

B

$h/g$

C

$h/2g$

D

$h/4g$

Solution

$\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=2 \mathrm{u} / \mathrm{g}$

$\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}$

$=\frac{g}{2}\left(t_{1}+t_{2}\right) t_{1}-\frac{1}{2} g t_{1}^{2}=\frac{1}{2} g t_{1} t_{2}$

$\therefore \mathrm{t}_{1}{t}_{2}=\frac{2 \mathrm{h}}{\mathrm{g}}$

Standard 11

Physics

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