JEE Mains · Physics · STD 11 - 3.2 motion in plane
The initial speed of a projectile fired from ground is \(u\). At the highest point during its motion, the speed of projectile is \(\frac{\sqrt{3}}{2} u\). The time of flight of the projectile is:
- A \(\frac{ u }{2 g }\)
- B \(\frac{ u }{ g }\)
- C \(\frac{2 u }{ g }\)
- D \(\frac{\sqrt{3} u }{ g }\)
Answer & Solution
Correct Answer
(B) \(\frac{ u }{ g }\)
Step-by-step Solution
Detailed explanation
\(u \cos \theta=\frac{\sqrt{3} u }{2} \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\) \(\Rightarrow \theta=30^{\circ}\) \(T =\frac{2 u \sin 30^{\circ}}{ g }=\frac{ u }{ g }\)
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