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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/$s^2$)
A
60 m/s
B
75 m/s
C
55 m/s
D
40 m/s
Detailed Solution
The first ball falls for 18 s from rest: $s_1 = \frac{1}{2}g(18)^2 = \frac{1}{2}\times10\times324 = 1620$ m
The second ball is thrown at t = 6 s, so it falls for 18 − 6 = 12 s: $s_2 = v\times12 + \frac{1}{2}g(12)^2 = 12v + 720$
They meet, so both have fallen the same distance: $12v + 720 = 1620$
$12v = 900$
v = 75 m/s
The second ball is thrown at t = 6 s, so it falls for 18 − 6 = 12 s: $s_2 = v\times12 + \frac{1}{2}g(12)^2 = 12v + 720$
They meet, so both have fallen the same distance: $12v + 720 = 1620$
$12v = 900$
v = 75 m/s