A ball is thrown vertically upwards. It was observed at a height h twice, with a gap of time interval `Delta t`. The initial velocity of the ball is :

Options

A. `sqrt(8gh+(g Deltat)^(2))`
B. `sqrt(2gh+((gDeltat)/(2))^(2))`
C. `sqrt(8gh+(2gDeltat)^(2))`
D. `sqrt(2gh)`

Answer

B

Explanation

Let the ball be at height h at time't' then
`h=ut-(1)/(2)gt^(2)rArrt^(2)-(2u)/(g)t+(2h)/(g)=0`
which has two roots `t_(1) & t_(2)rArr t_(1)t(2)=(2h)/(g)&t_(1)+t_(2)=(2u)/(g)`
But `t_(2)-t_(1)=Deltat` so `(Deltat)^(2)=(t_(1)+t_(2))^(2)-4t_(1)t_(2)`