JEE Mains · Physics · STD 11 - 3.2 motion in plane
A body of mass \(M\) thrown horizontally with velocity \(v\) from the top of the tower of height \(\mathrm{H}\) touches the ground at a distance of \(100 \mathrm{~m}\) from the foot of the tower. A body of mass \(2 \mathrm{M}\) thrown at a velocity \(\frac{v}{2}\) from the top of the tower of height \(4 \mathrm{H}\) will touch the ground at a distance of _______.
- A \(100\)
- B \(199\)
- C \(198\)
- D \(197\)
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
\(100=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} ; \quad \mathrm{x}=\frac{v}{2} \sqrt{\frac{2(4 \mathrm{H})}{\mathrm{g}}}=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(\Rightarrow \mathrm{x}=100\)
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