A ball (A) is thrown straight upfrom the edge the roof of a building. Another ball (B) is dropped from the of `1.00 s` later. You may ignore air resistance . (a) If the height of the building id `20.0 m`, what must the initial speed of ball (A) he if bothe are to hit the ground at the same time? (b) On the function of time, measured fro when the first ball is thrown and take origin at ground.
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Explanation

Let ball `A` is thrown upward at `t=0`. It takes time `T` to reach ground. Ball `B` is dripped from the roof `1.00 s` later. If both the balls reach the ground simultaneously, the flying time for ball `B` will be `T-ls`.
The displacement of both the balls is `(=-20 m)`. Using `y=y_(0)+ut +(1)/(2) at^(2)` initially for bath balls are at `y_(0)=20`
and when reaches ground where `y=0`.
For ball A: ` 0=20+v_(0)T+q(1)/(2)(-10)T^(2)` ..(i)
For ball B:`0=20+(1)/(2)(-10)(T-1)^(2)` ....(ii)
From Eqs. (i) and (ii),
`v_(0)T=5(2T-1)`
Frome Eqs. (i) and (ii) we get
`T^(2)-2T-3=0 rArr (t+1)(T-3)=0`
`t=3 s`
From Eqs. (iii) we get `v_(0)=(25)/(3) m s^(-1)`
Let ball `A` maximum height at time `t'` after throwing.
For ball `A` at maximum height, the velocity willbe zero.
Using `v=u+at`
`0=(25)/(3)-10xxt rArrt =(5)/(6) s`
for maximum height reached by ball `A`,
Using `v^(2)=u^(2)+2as`
`0=((25)/(3))^(2)-2xx10xxh rArr h =(125)/(36)m`
Hence, maximum height reach by ball `A`
`H=20+(125)/(36)=(845)/(36) m`.

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