Question

a parachutist drops freely from an aeroplane for 10 sec before the parachute opens out.then he descends with a net retardation of 25m/s2.if he bails out of the plane at a height of 2495 m and g = 10 m/s2, his velocity on reaching the ground will be..

Answer 1

Answer on Question #52582-Physics-Mechanics-Kinematics-Dynamics

A parachutist drops freely from an airplane for 10 sec before the parachute opens out. Then he descends with a net retardation of $25\mathrm{m/s^2}$. If he bails out of the plane at a height of $2495\mathrm{m}$ and $\mathrm{g} = 10\mathrm{m/s^2}$, his velocity on reaching the ground will be...

Solution

The velocity of the parachutist at the end of 10 seconds is $10g = 100\frac{m}{s}$ and the distance fallen in 10 seconds is $\frac{v^2}{2g} = \frac{100^2}{2 \cdot 10} = 500m$. The distance travelled after he bails out is

$$s = 2495 - 500 = 1995m.$$

For this distance $u = 100\frac{m}{s}$ and $a = -2.5\frac{m}{s^2}$. Therefore, the final velocity $v$ is given by

$$v^2 - u^2 = 2as;$$ $$v^2 = u^2 + 2as = 100^2 - 2 \cdot 2.5 \cdot 1995$$

Which gives $v = 5\frac{m}{s}$.

Answer: $v = 5\frac{m}{s}$.

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