A parachutist after bailing out falls 20,,m without friction. When parachute opens, it decelerates a

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2.Motion in Straight Line

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A parachutist after bailing out falls $20\,\,m$ without friction. When parachute opens, it decelerates at $2\,\,m/s^2.$ He reached the ground with a speed with a speed of $4\,\,m/s.$ At what height, did he bail out ?......$m$

A$91$

B$182$

C$293$

D$116$

Solution

$\mathrm{h}=\frac{\mathrm{u}^{2}-\mathrm{v}^{2}}{2 \mathrm{a}}$
$\mathrm{h}=\frac{400-16}{4}=\frac{384}{4}$
$h=96 \mathrm{m}$
$\mathrm{H}=20+96=116 \mathrm{m}$

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Solution

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