Question

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.

Answer 1

Correct Answer - A::B
(a) Time of descent `t = (sqrt(2H)/g) = (sqrt(2xx400)/10)`
`=8.94 s `
Now `v_x = ay = (sqrt5)y `
or `(dx)/(dt) = (sqrt5)(1/2g t^2) = 5(sqrt5)t^2`
`:. int_(0)^(x) dx = 5(sqrt5) int _(0)^t t^2 dt `
or horizontal drift
`x = (5sqrt(5))/3 (8.94)^3 = 2663m ~~ 2.67 km.`
(b) When particle strikes the ground
`v_x = sqrt(5)y = (sqrt5)(400) = 400(sqrt5)m//s`
`v_y = g t = 89.4 m//s`
`:. Speed = (sqrt(v_(x)^(2) + v_(y)^(2))) = 899m//s ~~ 0.9 km//s`.