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 ball meet at t = 18s. What is…

Question

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 ball meet at t = 18s. What is the value of v? [take g = 10 ms^–2]

Accepted Answer

Correct answer: A. (a) Let the two balls meet after ts at distance x from the platform. For the first ball u=0 ,t=18s ,g= 10 ms^2 Usingh= ut+ 12gt^2 ∴x= 12 10 18^2 … (i ) For the second ball u=v ,t=12 s ,g= 10 ms^2 Usingh= ut+ 12gt^2 ∴x=v 12+ 12 10 12^2 … From equations (i) and (ii), we get 12 10 18^2=12v+ 12 10 (12 )^2 or12v= 12 10 ( (18 )^2- (12 )^2)= 12 10 [ (18+12 )- (18-12 ) ]12v= 12 10 30 6 orv= 1 10 30 62 12=75 m/s

$A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two ball meet at t = 18s. What is the value of v? [take g = 10 m s^{-2}]$

Solution

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