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)
- 75 m/s
- 55 m/s
- 40 m/s
- 60 m/s
Answer
(1) 75 m/s
Solution
Distance fallen by the first ball in 18 s:
s1 = 12gt2 = 12 10 182 = 1620 m
The second ball moves for 18 - 6 = 12 s and covers the same distance:
s2 = v 12 + 12 10 122 = 12v + 720
s1 = s2
12v + 720 = 1620
v = 90012 = 75 m/s
Answer: (1)