TS EAMCET · Physics · Motion In Two Dimensions
Initial velocity with which a body is projected is \(10 \mathrm{~m} / \mathrm{s}\) from the base of an inclined plane as shown in the given figure. If the angle of projection is \(60^{\circ}\) with the horizontal, then the range \(R\) is [take, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ] 
- A \(\frac{15 \sqrt{3}}{2} \mathrm{~m}\)
- B \(\frac{40}{3} m\)
- C \(5 \sqrt{3} \mathrm{~m}\)
- D \(\frac{20}{3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(\frac{20}{3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Initial velocity of projectile, \(u=10 \mathrm{~m} / \mathrm{s}\) Angle of projection from inclined plane, \(\theta=60^{\circ}-30^{\circ}=30^{\circ}\) Here, \(\alpha=30^{\circ}\) Range over an inclined plane is given as…
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