A parachutist drops first freely from an aeroplane for $10 \mathrm{~s}$ and then parachute opens out. Now he descends with a net retardation of $2.5 \mathrm{~m} / \mathrm{s}^{2}$. If he bails out of the plane at a height of $2495 \mathrm{~m}$ and $g=10 \mathrm{~m} / \mathrm{s}^{2}$, his…

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

Aicharya · Accepted Answer

The velocity $v$ acquired by the parachutist after $10 \mathrm{~s}$ : $v=u+g t=0+10 \times 10=100 \mathrm{~m} / \mathrm{s}$ Then, $s_{1}=u t+\frac{1}{2} g t^{2}=0+\frac{1}{2} \times 10 \times 10^{2}=500 \mathrm{~m}$ The distance travelled by the parachutist under retardation, $s_{2}=2495-500=1995 \mathrm{~m}$ Let $v_{g}$ be this velocity on reaching the ground. Then $v_{g}^{2}-v^{2}=2 a s_{2}$ or $v_{g}^{2}-(100)^{2}=2 \times(-2.5) \times 1995$ or $v_{g}=5 \mathrm{~m} / \mathrm{s}$

A parachutist drops first freely from an aeroplane for \(10 \mathrm{~s}\) and then parachute opens out. Now…

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