Exams › NEET › Physics

A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (take g = 10 m/s²)

1. 75 m/s
2. 55 m/s
3. 40 m/s
4. 60 m/s

Correct answer: 75 m/s

Solution

The first ball falls freely under gravity, so its displacement after 18 s is s₁ = (1/2)gt² = 0.5 × 10 × (18)² = 1620 m. The second ball is thrown after 6 s, so it falls for 12 s with an initial velocity v. Its displacement is s₂ = vt + (1/2)gt² = v(12) + 0.5 × 10 × (12)² = 12v + 720. Since they meet, s₁ = s₂, so 1620 = 12v + 720. Solving, v = 75 m/s.

Related NEET Physics questions

• In the equation of motion, \( \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ \) \( 1 / 2 a t^{2}, \) S stands for
• A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?
• A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is
• From the displacement-time graph shown here, find the velocity of the body as it moves from \( A \) to \( B \)
• The distance of a particle as a function of time is shown below. The graph indicates that
• Whenever an object moves with a constant speed, its distance - time graph is a

⚔️ Practice NEET Physics free + battle 1v1 →