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NEET PHYSICSEasy

A ball is projected with a velocity of 10 m/s10 \text{ m/s} at an angle of 60∘60^{\circ} with the vertical direction. Its speed at the highest point of its trajectory will be:

A

10 m/s10 \text{ m/s}

B

Zero

C

53 m/s5\sqrt{3} \text{ m/s}

D

5 m/s5 \text{ m/s}

Step-by-Step Solution

  1. Analyze the Geometry: The ball is projected at an angle of 60∘60^{\circ} with the vertical. Therefore, the angle of projection θ\theta with the horizontal is: θ=90∘−60∘=30∘\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}
  2. Analyze Velocity at Highest Point: In projectile motion, the horizontal component of velocity (uxu_x) remains constant throughout the flight because there is no acceleration in the horizontal direction (neglecting air resistance). At the highest point of the trajectory, the vertical component of velocity (vyv_y) becomes zero. Thus, the total speed at the highest point is equal to the horizontal component of the initial velocity.
  3. Calculate Speed: vhighest=ux=ucos⁡θv_{\text{highest}} = u_x = u \cos \theta Given u=10 m/su = 10 \text{ m/s} and θ=30∘\theta = 30^{\circ}: vhighest=10cos⁡(30∘)v_{\text{highest}} = 10 \cos(30^{\circ}) vhighest=10×32=53 m/sv_{\text{highest}} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m/s}
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