A particle is projected with a velocity of `30 m//s` at an angle `60^(@)` above the horizontal on a slope of inclination `30^(@)`. Find its time of flight.

Explanation

The coordinate system, projection velocity and its component, and acceleration due to gravity and its component are shown in the adjoining figure. Subsituting corresponding values in following equation, we get the time of flight.
`T=(2u_(y))/a_(y) rarr T=(2xx15)/(5sqrt(3))=2sqrt(3) s`

Substituting value of time of flight in following equation, we get the range R.
`R=u_(x)T-1/2a_(x)T^(2)rArr R=15sqrt(3)xx2sqrt(3)-1/2xx5xx(2sqrt(3))^(2)=60 m`
In the adjoining figure, components of velocity `vec(v)_(P)` when the projectile hits the slope at point P are shown. The angle `beta` which velocity vector makes with the x-axis is known as angle of hit. The projectile hits the slope with such a velocity `vec(v)_(P)`, whose y-component is equal in magnitude to that of velocity of projection. The x-component of velocity `v_(x)` is calculated by substituting value of time of flight in following equation.
`v_(x)=u_(x)-a_(x)t rarr v_(x)=15sqrt(3)-5xx2sqrt(3)=5sqrt(3)`
`beta=tan^(-1) (v_(y)/v_(x)) rarr beta =60^(@)`