JEE Mains · Physics · STD 11 - 2. motion in straight line
A balloon was moving upwards with a uniform velocity of \(10\, {m} / {s}\). An object of finite mass is dropped from the balloon when it was at a height of \(75\, {m}\) from the ground level. The height of the balloon from the ground when object strikes the ground was around.(In \({m}\)) (Takes the value of \(g\) as \(10 \,{m} / {s}^{2}\) )
- A \(125\)
- B \(250\)
- C \(300\)
- D \(200\)
Answer & Solution
Correct Answer
(A) \(125\)
Step-by-step Solution
Detailed explanation
\(S=u t+\frac{1}{2} a t^2\) \(-75=+10 t+\frac{1}{2}(-10) t^2\) \(\Rightarrow t=5 sec\) Object takes \(t=5 s\) to fall on ground Height of balloon from ground \(H=75+u t\) \(H=75+10 \times 5=125 m\)
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