The windshield of a truck is inclined at `37^(@)` to the horizontal. The truck is moving horizontally with a constant acceleration of `a = 5 m//s^(2)`. At the instant the velocity of the truck is `v_(0) = 0.77 m//s`, an insect jumps from point A on the windshield, with a velocity `u = 2.64 m//s` (relative to ground) in vertically upward direction. It falls back at point B on the windshield. Calculate distance AB. Assume that the insect moves freely under gravity and `g = 10 m//s^(2)`.

Answer

`AB = 0.57 m`