Question:

A balloon rises from ground with an acceleration of $1.25\, m s^{-2}$. After $8s$ a stone is released from balloon. The stone will

Updated On: Apr 18, 2024
  • cover a distance of 45 m in reaching the ground
  • have a displacement of 50 m
  • begin to move down after being released
  • reach the ground in 4 s
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The Correct Option is D

Solution and Explanation

S = u + $\frac{1}{2} at^2 = \frac{1}{2} \times 1.25 \times 8^2 $ = 40 m
Using V = u + at we get $\nu$ = 1.25 $\times$ 8 = 10 m $s^{-1}$
Using S = ut + $\frac{1}{2} at^2$ for dropped stone, we get
40 = -10t $\times + \frac{1}{2} \times 10t^2$ i.e. $5t^2$ - 10t - 40 = 0 i.e. t = 4 s or -2 s
Taking the possible value t = 4 s .
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