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Standard XII

Physics

Dropping Objects

A parachutist...

Question

A parachutist drops first freely from an aeroplane for $10 s$ and then his parachute opens out. Now 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 $g = 10 m s^{- 2}$ , his velocity on reaching the ground will be

5 m s^{- 1}

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10 m s^{- 1}

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15 m s^{- 1}

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20 m s^{- 1}

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Solution

The correct option is D $5 m s^{- 1}$
Initial velocity $(u) = 0$
From $t = 0$ to $t = 10$ free fall
$V = 0 + (10) (10) V = 100 m . s$
Distance covered $= + \frac{1}{2} \times 10 \times 10^{2} = 500 m$
After this retardation of $a = 2.5 m / s^{2}$
Now from here tall it reaches ground-
Distance covered $=$ total distance $-$ the distance covered in first $10 s e c$
$= (2495 - 500) m = 1995 m$
Now, By $3^{r d}$ equation of motion-
$v^{2} = u^{2} + 2 a s v^{2} = (100)^{2} + 2 (- 2.5) (1995) = 10000 - 9975 = 25 \Rightarrow v = 5 m / s$

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

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