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. `2.67 km`
B. `5.67 km`
C. `12.67 km`
D. `4.97 km`

Answer

A

Explanation

`v_(y)=(dy)/(dt)=sqrt(2gy)`...(1)
`v_(x)=(dx)/(dt)=ay`...(2)
Dividing (1) by (2), we get
`(dy)/(dx)=sqrt(2gy)/ay=2/sqrty`
`(g=10m//s^(2) and a=sqrt5 s^(-1))`
or `sqrty.dy=2dx`
or `int_(0)^(400 m)sqrty.dy=2 dx`
or `2/3(y^(3//2))_(0)^(400 m)=2x`
or `x=1/3(400)^(3//2)`
or `x=8/3xx10^(3)m`
or `x=2.67 km`