A particle Is projected at point `'A'` with initial velocity `5 m//s` at an angle `theta = 37^(@)` with the vertical `y` axis. A strong horizontal wind gives the particle a constant horizontal acceleration `6 m//s^(2)` in the `x` direction. If the particle strikes the ground at a point directly under its released position, determine the height `'h'` of point `A`. The downward acceleration is taken as the constant
`g = 10 m//s^(2)`(take `sin37^(@) = (3)/(5), cos37^(@) = (4)/(5)`)

Answer

`9`

Explanation

For `A` to `B`

`s_(x) = u_(x)t + (1)/(2) a_(x)t^(2)`
`0 = - usintheta t + (1)/(2) at^(2)`
`t = (2usintheta)/(a) = (2(5)sin37^(@))/(6) = 1` sec.
`s_(y) = h = AB = u_(y)t + (1)/(2) a_(y)t^(2) = ucostheta t + (1)/(2)g t^(2)`
`= (5)((4)/(5))(1) + (1)/(2)(10)(1)^(2) = 9m`