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Q.
A balloon rises from ground with an acceleration of $2\, m / s ^{2}$. After $10$ second, a stone is released from the balloon. The stone will
Motion in a Straight Line
Solution:
After $t=10\, \sec$
$v _{1} = u + at$
$=0+2 \times 10$
$v _{1} =20\, m / \sec$
From A to B
$H = ut +\frac{1}{2} at ^{2}$
$=0 \times 10+\frac{1}{2} \times 2 \times 100$
$H =100\, m$
At maximum height
$v^{2}=v_{1}^{2}+2 a h$
$0=(20)^{2}-2 \times 10 \times h$
$h=\frac{400}{20}=20\, m$
So, total distance is $BC + CB + BA$
$=20+20+100=140\, m$
