A body of mass m is projected horizontally with a velocity v from the top of a tower of height h and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2m is projected horizontally from the top of a tower of height 2h, it reaches the ground at a distance 2x from the foot of the tower. The horizontal velocity of the second body is :

Explanation

For the first body, `h = 1/2 " gt"^(2) " "...(i) `
and ` a = "vt " ...(ii) `
From equations (i) and (ii), we get
`h=1/2g(x^(2)/v^2)" "... (iii)`
For the second body, let v. be the velocity of projection, then
`2h=1/2g[((2x)^2)/(v.^(2))]" "....(iv)`
Dividing equations (iii) by equation (iv), we get
`1/2=x^2/y^2xx(v.^2)/(4x^2)" or " v.=sqrt(2v)`