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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

With what velocity a ball be projected vertically so that the distance covered by it in 5th second is twice the distance it covers in its 6th second (g=10 m/s2g=10 \text{ m/s}^2)?

A

58.8 m/s

B

49 m/s

C

65 m/s

D

19.6 m/s

Step-by-Step Solution

  1. Formula for Distance in n-th Second: The distance covered by a particle in the nn-th second of uniformly accelerated motion is given by Sn=u+a2(2n1)S_n = u + \frac{a}{2}(2n - 1) . For vertical upward motion, a=g=10 m/s2a = -g = -10 \text{ m/s}^2.
  2. Calculate Distances:
  • Distance in 5th second (n=5n=5): S5=u102(2×51)=u5(9)=u45S_5 = u - \frac{10}{2}(2 \times 5 - 1) = u - 5(9) = u - 45
  • Distance in 6th second (n=6n=6): S6=u102(2×61)=u5(11)=u55S_6 = u - \frac{10}{2}(2 \times 6 - 1) = u - 5(11) = u - 55
  1. Apply Condition: The problem states S5=2S6S_5 = 2 S_6. u45=2(u55)u - 45 = 2(u - 55) u45=2u110u - 45 = 2u - 110 u=11045=65 m/su = 110 - 45 = 65 \text{ m/s}
  2. Verification of Direction: The time to reach maximum height is tmax=u/g=65/10=6.5 st_{max} = u/g = 65/10 = 6.5 \text{ s}. Since both the 5th second (t=4t=4 to 55) and 6th second (t=5t=5 to 66) occur before tmaxt_{max}, the ball is moving purely upwards during these intervals, so distance equals displacement magnitude.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEvelocityprojectedverticallydistancecovered

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