A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take `g = 10 m//s^2.`

Options

A. 45 m
B. 50 m
C. 60 m
D. 65 m

Answer

A

Explanation

Velocity of `1 st` stone when passing at `A`
`V^2 = 0 + 2.10.5 rArr V = 10 m//s`
And `S_1 - S_2 = 20 m`
`rArr (10. t + (1)/(2) 10.t^2) - ((1)/(2) . 10.t^2) = 20`
At `t = 2 s, S_2 = (1)/(2) g t^2 = (1)/(2) xx 10 xx 2^2 = 20 m`
Hence height of the tower,
`H = S_1 + S_2 = 25 + 20 = 45 m`.