MHT CET · Physics · Motion In Two Dimensions
A cricket player hit a ball like a projectile, but the fielder caught the ball after 2 second. The maximum height reached by the ball is
\(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(2 \mathrm{~m}\)
- B \(5 \mathrm{~m}\)
- C \(4 \mathrm{~m}\)
- D \(3 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Time of flight, \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=2 \mathrm{~s}\)
\(\therefore \mathrm{u} \sin \theta=\mathrm{g}=10\)
Maximum height, \(\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{(10)^2}{2 \times 10}=5 \mathrm{~m}\)
\(\therefore \mathrm{u} \sin \theta=\mathrm{g}=10\)
Maximum height, \(\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{(10)^2}{2 \times 10}=5 \mathrm{~m}\)
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