Question

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 the second body of mass of 4 m is projected horizontally from the top of a tower of the height of 4 h, it reaches the ground at a distance of 4x from the foot of the tower. The horizontal velocity of the second body is:

पर्याय

• 6V

• `sqrt2"V"`

• 2V

• 5V

Answer 1

2V

Explanation:

For the first body h = `1/2 "gt"^2` ...(u = 0)

t = `"x"/"v"`

∴ h = `1/2 "g"("x"^2/"v"^2)` ....(1)

`"h" ∝ "x"^2/"v"^2`

∴ `"h"/(4"h") = "x"^2/"v"^2 xx "v'"^2/(4"x")^2`

4v² = v'²

2v = v'