Question

standing on the roof,you simultaneously throw one ball straight up and drop another from rest .Which hits the ground at first?

Answer 1

let hight of roof = H
for 1st ball ,
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when ball thrown upward , then,
let initial velocity = u
use formula,
-h = ut- 1/2gt²
-2h = 2ut -gt²
gt² -2ut - 2h =0
t = { 2u ±√(4u²+8gh)}/2g

={ u ±√( u² +2gh)}/g
t≠ negative so,

t ={ u +√(u² +2gh)}/g
=u/g + √( u²/g² +2h/g)

for 2nd ball
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time taken by ball let T
S = ut +1/2at²
here u =0
h = 1/2gT²
T² = 2h/g
T = √(2h/g)

here we see t > T
so, second ball which drop form the roof reaches ground first