question_answer

A balloon starting from the ground has been ascending vertically at a uniform velocity for 4.5 s and a stone left fall from it reaches the ground in 7 s. The velocity of the balloon when the stone was left fall is \[\left[ use\text{ }g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}} \right].\]

A) 20.88 m/s                     

B) 20 m/s

C) 22 m/s              

D) 24 m/s

Correct Answer: A

Solution :

let \[{{v}_{0}}\]=the constant velocity with which the balloon is going up. It will also be the initial velocity of the stone dropped from it. Let us consider the point from where the stone is dropped as origin.

For balloon; let \[h=\]height of the balloon when the stone is dropped.

\[h=4.5{{v}_{0}}..(i)\]

For the stone \[y={{v}_{0}}t+1/2a{{t}^{2}}\]

\[\Rightarrow -h={{v}_{0}}t+1/2\left( -g \right)\times {{t}^{2}}\]

\[~\left[ \operatorname{coordinate} of ground =-h \right]\]

\[\Rightarrow -4.5{{v}_{0}}={{v}_{0}}\times 7-1/2\times 9.8\times {{\left( 7 \right)}^{2}}\]\[\Rightarrow {{v}_{0}}=20.88m/s\]