COMEDK · Physics · 4. Motion In Two Dimensions
A piece of marble is projected from Earth's surface with velocity of \(19.6 \sqrt{2} \mathrm{~m} / \mathrm{s}\) at \(45^{\circ}\). After 2 s , its velocity makes an angle \(\alpha\) with horizontal, where \(\alpha\) is
- A \(45^{\circ}\)
- B \(30^{\circ}\)
- C \(60^{\circ}\)
- D \(0^{\circ}\)
Answer & Solution
Correct Answer
(D) \(0^{\circ}\)
Step-by-step Solution
Detailed explanation
\(u_x=19.6 \mathrm{~m} / \mathrm{s}\) and \(u_y=19.6 \mathrm{~m} / \mathrm{s}\). After 2 s , vertical component of velocity \(v_y\) will become zero. So, particle is at its highest point.
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