First slide

Rectilinear Motion

Question

A parachutist drops freely from an airplane for 10 s before the parachute opens. He then descends with a uniform retardation of 2.5 ms^-2. If he bails out of the plane at a height of 2495 m and g is 10 ms^-2, his velocity on reaching the ground will be

Moderate

A

5 ms^-1
B

10 ms^-1
C

15 ms^-1
D

20 ms^-1

Solution

Initial velocity of dropping = zero
Let v₁ be velocity at end of 10 s
$\begin{array}{l} \Rightarrow v_{1} = gt \\ \Rightarrow v_{1} = 100 {ms}^{- 1} \end{array}$
Distance travelled during this time is
$\begin{array}{l} h_{1} = \frac{v_{1}^{2}}{2 g} = \frac{(100)^{2}}{2 (10)} \\ \Rightarrow h_{1} = 500 m \end{array}$
So, a remaining distance of 2495-500 = 1995 m has to be travelled with a retardation of 2.5 ms^-2. Let the parachutist strike the ground with velocity v.
Then
$\begin{array}{l} v^{2} - {v_{1}}^{2} = 2 a (h - h^{'}) \\ \Rightarrow v^{2} - (100)^{2} = 2 (- 2 .5) (1995) \\ \Rightarrow v^{2} = 10000 - 9975 \\ \Rightarrow v^{2} = 25 \end{array} \Rightarrow v = 5 {ms}^{- 1}$

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