First slide

Release from certain height

Question

A parachutist after bailing out falls 50 m without friction when parachute opens, it decelerates at $2 {ms}^{- 2}$ . He reaches the ground with speed of 3 m/ s. At what height did he bail out?

Moderate

A

293 m
B

111 m
C

91 m
D

182 m

Solution

After bailing out from point A, parachutist falls freely under gravity. The velocity acquired by it will be v.
From $v^{2} = u^{2} + 2 as = 0 + 2 x 9.8 x 50 = 980$
(As, u = 0, a = 9.8 ${ms}^{- 2}$ , s = 50 m)
At point B, parachute opens and it moves with retardation of $2 {ms}^{- 2}$ and reach at ground (point C with
velocity of $3 {ms}^{- 1}$ ).
For the part BC by applying the equation $v^{2} = u^{2} + 2 as$
$v = 3 {ms}^{- 1}, u = \sqrt{980} {ms}^{- 1}, a = - 2 {ms}^{- 2}, s = h$
$\Rightarrow (3)^{2} = (\sqrt{980})^{2} + 2 \times (- 2) \times h$
$\Rightarrow 9 = 980 - 4 h$
$\Rightarrow h = \frac{980 - 9}{4} = \frac{971}{4} = 242.7 ≅ 243 m$
So, the total height by which parachutist bails out
= 50 + 243 = 293 m

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