Question

A balloon rises from ground with an acceleration of $2m/s^2$. After $10$ second, a stone is released from the balloon. The stone will

• A. have a displacement of $150m$
• B. cover a distance of $140m$ in reaching the ground after release
• C. reach the ground in 4 second
• D. begin to move downward after being released

Answer 1

Answer: (B)

After $t=10\sec$
$v_1=u+at$
$=0+2\times 10$
$v_1=20m/\sec$
From A to B
$H=ut+\frac{1}{2}a{t}^2$
$=0\times 10+\frac{1}{2}\times 2\times 100$
$H=100m$
At maximum height
$v^2=v_1^2+2ah$
$0=(20{)}^2-2\times 10\times h$
$h=\frac{400}{20}=20m$
So, total distance is $BC+CB+BA$
$=20+20+100=140m$