An object is dropped from an altitude of one Earth radius above Earth’s surface. If M is the mass of Earth and R is its radius. The speed of the object just before it hits Earth is given by:

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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