Kinematics Question 225
A balloon starts rising from the ground with an acceleration of 1.25 m/s2 after 8s, a stone is released from the balloon. The stone will ( $ g=10 $ m/s2) [KCET 2001]
Options:
A) Reach the ground in 4 seconds
B) Begin to move down after being released
C) Have a displacement of 50 m
D) Cover a distance of 40 m in reaching the ground
Show Answer
Answer:
Correct Answer: A
Solution:
When the stone is released from the balloon. it’s height $ h=\frac{1}{2}at^{2}=\frac{1}{2}\times 1.25\times {{(8)}^{2}}=40\ m $ and velocity $ v=at=1.25\times 8=10\ m/s $ Time taken by the stone to reach the ground $ t=\frac{v}{g}[ 1+\sqrt{1+\frac{2gh}{v^{2}}} ]=\frac{10}{10}[ 1+\sqrt{1+\frac{2\times 10\times 40}{{{(10)}^{2}}}} ] $ =4 sec
BETA
