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 :

Options

A. `v`
B. `2v`
C. `sqrt(2)v`
D. `v//2`

Answer

c

Explanation

For the first body, `h=1/2"gt"^(2)........(i)`
and `c=vt..........(ii)`
From eqn. (i) and(ii)
`h=1/2g.((x^(2))/(v^(2)))...........(iii)`
Dividing eqn (iii) by eqn (iv) we get
`1/2=(x^(2))/(v^(2))xx(v'^(2))/(4x^(2)) or v'=sqrt2v`