posted by Anonymous .
A ball is dropped from rest from a height of 20.0 m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?
What's the acceleration you're told to use in your class?
9.8m/s^2? 10m/s^2? 9.81m/s^2?
I'll go with 10 because I'm lazy, just substitute to whatever you guys are using.
The first ball (dropped from rest) will take 2 seconds to reach the ground :
20 = 5 x t^2 and you t=2
As such, the second ball will reach the ground in 1 second flat :
20 = Vinitial x t + 5 x t^2 where t=1
20 = Vinitial + 5
Answer : Vinitial = 15
Really, you just needed to use the equation you're given : x = x0 + v0*t + 1/2a*t^2