A particle is released from a certain height `H = 400m`. Due to the wind, the particle gathers the horizontal velocity component `v_x = ay "where a "= (sqrt5)s^(-1)` and y is the vertical displacement of the particle from the point of release, then find
(a) the horizontal drift of the particle when it strikes the ground ,
(b) the speed with which particle strikes the ground.

Options

A. 2.67km
B. 8.67km
C. 1.67km
D. 5.1km

Explanation

`V_(y)=(dy)/(dt)=sqrt(2gy)`. . . (1)
`V_(x)=(dx)/(dt)=ay` . . . .(2)
Dividing 1 by 2
`(dy)/(dx)=sqrt(2gy)/(ay)=(2)/sqrt(y)`
or `sqrt(y)dy=2dx`
Integrating
`int_(0)^(400)=2int_(0)^(x)dx`

`therefore x=2.67`