Question

Question #1524e

Answer 1

Let,the deceleration due to air resistance acting on the particle upwards , is `kx` (where, `k` is a constant and `x` is the height of the object above the ground)

So,net acceleration of the object is `g-kx` downwards

So,we can write, acceleration = `(dv)/(dt) = g-kx`

or, `(dv)/(dx) (dx)/(dt) = g-kx`

or, `v (dv)/(dx) = g-kx`

so, `vdv = gdx - kxdx`

so, `int_0^V vdv= int_0^H dx - k int_0^H xdx`(considering that on reaching the ground from `H` above the ground,it will have velocity of `V`)

So, `V^2/2 =gH - kH^2/2`

So, `V^2 = 2gH -kH^2`

This is the relationship derived for this body.