A parachutist drops freely from an aeroplane& | vedclass

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Question

The initial velocity of the parachutist is zero

now $\mathrm{V}=\mathrm{u}+\mathrm{gt}$

$\mathrm{V}=0+(-10) 10$

$\mathrm{V}=-100$( $-sign$ as

Vedclass · Accepted Answer

$5$

Vedclass · Answer

The initial velocity of the parachutist is zero

now $\mathrm{V}=\mathrm{u}+\mathrm{gt}$

$\mathrm{V}=0+(-10) 10$

$\mathrm{V}=-100$( $-sign$ as the velocity is in downward direction)

Now $\mathrm{S}=\mathrm{ut}+1 / 2 \mathrm{ut}^{\wedge} 2$

so $S=0+1 / 2(-10) 	imes100$

$\mathrm{S}=500$

Now $\mathrm{H}=2495-500$

$=1995$

NOw $v^2=u^ 2+2 a h$

$v^2=1000-2	imes 2.5	imes 1995$

$\mathrm{v}^2=25$

$v=5$

so his velocity on reaching the ground is $5 \mathrm{m} / \mathrm{s}$

A parachutist drops freely from an aeroplane for $10\,s$ before the parachute opens out. Then he descends with a net retardation of $2.5\, m/s^2$. If he bails out of the plane at a height of $2495\, m$ and $g = 10\, m/s^2$, hit velocity on reaching the ground will be .......$m/s$

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