WBJEE · Physics · Motion In Two Dimensions
A body is projected with a speed \(u \mathrm{~m} / \mathrm{s}\) at an angle \(\beta\) with the horizontal. The kinetic energy at the highest point is \(3 / 4\) th of the initial kinetic energy. The value of \(\beta\) is :
- A \(30^{\circ}\)
- B \(45^{\circ}\)
- C \(60^{\circ}\)
- D \(120^{\circ}\)
Answer & Solution
Correct Answer
(A) \(30^{\circ}\)
Step-by-step Solution
Detailed explanation
Hints: (K.E.) at maximum height \(=\frac{1}{2} m\left(u^2 \cos ^2 \beta\right)\) \(\mathrm{K} . \mathrm{E} .=\mathrm{K} \cos ^2 \beta\) Here, \(K \cos ^2 \beta=\frac{3}{4} \mathrm{~K}\) \[ \begin{aligned} & \cos \beta=\frac{\sqrt{3}}{2} \\ & \beta=30^{\circ} \end{aligned} \]
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