MHT CET · Physics · Motion In Two Dimensions
A body when projected at an angle ' \(\theta\) ' with the horizontal reaches a maximum height ' H '. The time of flight of the body will be ( \(g=\) acceleration due to gravity)
- A \(\frac{1}{2} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\)
- B \(\sqrt{\frac{g}{2 H}}\)
- C \(2 \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\)
- D \(\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\)
Step-by-step Solution
Detailed explanation
\( H = \frac{(u \sin \theta)^2}{2g} \) \( u \sin \theta = \sqrt{2gH} \)
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