JEE Main 2023 · 12 April · morning · Single correct · +4 / −1

A ball is thrown vertically upward with an initial velocity of 150 m / s. The ratio of velocity after 3 s and 5 s is (x+1)/x. The value of x is ___________.
{ take, g=10 m / s²}

  1. (a) -5
  2. (b) 10
  3. (c) 5
  4. (d) 6
Answer and solution

Answer (c) 5

Solution

Physics sir · 4 steps

Find both velocities with v = u − gt, then match their ratio to the form (x+1)/x.

  1. Velocity equation. Take upward as positive, so gravity is -g. The ball rises for 150/10 = 15 s, so at 3 s and at 5 s it is still moving up.

    v = u - gt

  2. Put in numbers. At t = 3 s: v₃ = 150 - 10 × 3 = 120 m s⁻¹. At t = 5 s: v₅ = 150 - 10 × 5 = 100 m s⁻¹.
  3. Match the form. v₃/v₅ = 120/100 = 6/5. Writing this as (x+1)/x needs a denominator one less than the numerator, so x = 5.
  4. So. x = 5, which is option (c).

The idea to keep. In vertical motion, fix one direction as positive and use v = u - gt. Check that the ball has not turned around before the given time.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Rectilinear Motion

Asked about 110 times in JEE Main since 2020. This question is from 2023.