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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, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]

  1. A \(\frac{15 \sqrt{3}}{2} \mathrm{~m}\)
  2. B \(\frac{40}{3} m\)
  3. C \(5 \sqrt{3} \mathrm{~m}\)
  4. D \(\frac{20}{3} \mathrm{~m}\)
Verified Solution

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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