A ball (A) is thrown straight up from the edge of the roof of a building. Another ball (B) is dropped from the roof `1.00 s` later. You may ignore air resistance . (a) If the height of the building is `20.0 m`, what must the initial speed of ball A be if both are to hit the ground at the same time? (b) On the same graph sketch the position and velocity of each balls a function of time, measured from when the first ball is thrown and taken origin at ground.
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Explanation

For the purpose of doing all four farts with the least reperition of algebra, quantities will be dented symbolically. That is, let
`y_(1)=h+v_(0)t-(1)/(2)g t^(1)` and `y_(2)=h-(1)/(2)g(t-t_(0))^(2)`.
In this case, `t_(0)=1.00 s`. Setting `y_(1)=y_(2)=0`, expanding the binomial `(t-t_(0))^(2)`, which can be solved for `t`,
`t=((1)/(2)g t_(0)^(2))/(g t_(0)-v_(0)) =t_(0)/(2) (1)/((1-v_(0)/(g t_(0)))` Substiuting this into the expressinon for `y_(1)` and setting `y_(1)=0` and solving for `h` as a function of `v_(0)` yields after some algebra,
`h=(1)/(2) g t_(0)^(2)((1)/(2)g t_(0)-v_(0))^(2)/(g t_(0)-v_(0))^(2)`
a. Using the given value, `t_(0)=1.00 s` and `g=9.80 ms^(-2)`.
`h=20.0 m=(4.9 m) ((4.9 ms^(-1)-v_(0))/(9.8 ms^(-2)-v_(0)))`
This has two solutions, one of which is not physical [the first ball is still going up when the second is released, see part `C `]. The physical solution invlves taking the negative square root before solving for `v_(0)`, and yields `8.2 ms^(-1)`
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b. The above expression gives
for (i) `0.411 m` and for (ii) `1.15 km`.
C. As `v_(0)` approaches `9.8 ms^(-1)`, the height `h` becomes infinite corresponding to a relative velocity at the second ball is throun that approaches zero, If `v_(0)le 9.8 ms^(-1)`, the first ball can never catch the second ball.
d. As `v_(0)` approaches `4.9 ms^(-1)` the height approaches zero. This corresponds to the first ball being closer and is released. If `v_(0) le 4.9 ms^(-1)`, the first ball will already have roof on the wau down nefore the second ball is released, and the second ball can catch up.