Hard
Question
A ball is dropeed from a high rise plateform at t = 0 starting from rest. After 6 second another ball is thrown downwards from the same plateform with a speed v. The ball meet at t = 18 s. What is the value of v ?
Options
A.75 m/s
B.55 m/s
C.40 m/s
D.60 m/s
Solution
Let the two balls meet after t s at distance x from the plateform For the first balls
u = 0, t = 18s, g = 10 m/s2
Using h = ut +
gt2
∴ x = × 10 × 182 ......(i)
For the secons balls
u = v, t = 12s, g = 10 m/s2
Using h = ut + gt2
∴ x = v × 12 +× 10 × 122 ......(ii)
From equation (i) and (ii), we get
× 10 × 182 = 12 v + × 10 × 122
or 12 v = × 10 × [182 - 122]
12 v = × 10 × [(18 + 12)(18 - 12)]
12 v = × 10 × 30 × 6
or v =
= 75 m/s.
u = 0, t = 18s, g = 10 m/s2
Using h = ut +
gt2∴ x =
× 10 × 182 ......(i)For the secons balls
u = v, t = 12s, g = 10 m/s2
Using h = ut +
gt2 ∴ x = v × 12 +
× 10 × 122 ......(ii)From equation (i) and (ii), we get
× 10 × 182 = 12 v + × 10 × 122or 12 v =
× 10 × [182 - 122]12 v =
× 10 × [(18 + 12)(18 - 12)]12 v =
× 10 × 30 × 6or v =
= 75 m/s.Create a free account to view solution
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