MHT CET · Physics · Motion In Two Dimensions
A shell is fired at an angle of \(30^{\circ}\) to the horizontal with velocity \(196 \mathrm{~m} / \mathrm{s}\). The time of flight is \(\left[\sin 30^{\circ}=\frac{1}{2}=\cos 60^{\circ}\right]\)
- A \(10 \mathrm{~s}\)
- B \(16.5 \mathrm{~s}\)
- C \(20 \mathrm{~s}\)
- D \(6.5 \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(20 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
We know the time of flight for a projectile is given by:
\(t=\frac{2 u \sin \theta}{g}\)
On placing given values in the above equation:
\(t=\frac{20 \times(196 \mathrm{~m} / \mathrm{s}) \times \sin \left(30^{\circ}\right)}{\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)}=20 \mathrm{~s}\)
\(t=\frac{2 u \sin \theta}{g}\)
On placing given values in the above equation:
\(t=\frac{20 \times(196 \mathrm{~m} / \mathrm{s}) \times \sin \left(30^{\circ}\right)}{\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)}=20 \mathrm{~s}\)
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