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 \[y={{v}_{0}}t+1/2a{{t}^{2}}\] |
\[\Rightarrow -h={{v}_{0}}t+1/2\left( -g \right)\times {{t}^{2}}\] |
\[~\left[ \operatorname{coordinate} of ground =-h \right]\] |
\[\Rightarrow -4.5{{v}_{0}}={{v}_{0}}\times 7-1/2\times 9.8\times {{\left( 7 \right)}^{2}}\]\[\Rightarrow {{v}_{0}}=20.88m/s\] |
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