JEE Mains · Physics · STD 11 - 3.2 motion in plane
A body is projected from the ground at an angle of \(45^{\circ}\) with the horizontal. Its velocity after \(2s\) is \(20 \,ms ^{-1}\). The maximum height reached by the body during its motion is \(m\). (use \(g =10\, ms ^{-2}\) )
- A \(20\)
- B \(25\)
- C \(29\)
- D \(200\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
\(v \cos \alpha=u \cos 45^{\circ}\) \(v \sin \alpha=u \sin 45^{\circ}-g t\) Solve for \(u\) we get \(u=20 \sqrt{2} m /s\) \(\Rightarrow H=\frac{u^{2} \sin ^{2} 45^{a}}{20}=20 m\)
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