-
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 = 18s. What is the value of v?[2010] (take g = 10 m/s²)
-
- 75 m/s
- 55 m/s
- 40 m/s
- 60 m/s
Correct Option: A
Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s.
Now, distance moved in 18 s by 1st ball = 1/2 × 10 × 18² = 90 × 18 = 1620 m
Distance moved in 12 s by 2nd ball
= ut + gt²
∴ 1620 = 12 v + 5 × 144
⇒ v = 135 – 60 = 75 ms–1
