First slide

Rectilinear Motion

Question

A paruchutist 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 ms^-2. If he bails out of the plane at a height of 2495 m and g = 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

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

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