A balloon starts rising from the ground with an acceleration of $1.25 m s^{- 2}$ , After $8$ seconds, a stone is released from the balloon, The stone will
$(g = 10 m s^{- 2})$

cover a distance of 40 m in reaching the ground.

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have a displaceent of 50 m

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reach the ground in 4 second

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begin to move down after being released

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Solution

The correct option is D reach the ground in 4 second
$s = \frac{1}{2} a t^{2}$
$a = 1.25 m / s^{2} \Rightarrow s = \frac{1}{2} \times 1.25 \times 8^{2} = 40 m$
After $8 s e c$ speed of balloon:
$v = u - g t$
$v = - (1.25) \times 8 = - 10 m / s$
Now, when stone is released distance to cover is $40 m$ .
$v^{2} = u^{2} - 2 g s$
$v^{2} = (100)^{2} + 2 \times 10 \times 40 \Rightarrow v = 30 m / s$
$v = u - g t \Rightarrow - 30 = + 10 - 10 t$
$t = 4 s$

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A balloon starts rising from the ground with an acceleration of 1.25ms−2, After 8 seconds, a stone is released from the balloon, The stone will (g=10ms−2)

The correct option is D reach the ground in 4 seconds=12at2a=1.25 m/s2⇒s=12×1.25×82=40 mAfter 8 sec speed of balloon:v=u−gtv=−(1.25)×8=−10 m/sNow, when stone is released distance to cover is 40 m.v2=u2−2gsv2=(100)2+2×10×40⇒v=30m/sv=u−gt⇒−30=+10−10tt=4 s

A balloon starts rising from the ground with an acceleration of $1.25 m s^{- 2}$ , After $8$ seconds, a stone is released from the balloon, The stone will
$(g = 10 m s^{- 2})$