Let a particle be projected up with a speed v_{0} at an angle θ to horizontal onto an inclined plane of inclination β .

Hence the component of initial velocity (velocity of projection) parallel and perpendicular to the plane are equal to v_{0} cos (θ – β) and v_{0} sin (θ – β) respectively.

The component of g along the plane is gsinβ and perpendicular to the plane is gcosβ as shown in the Figure.

Therefore, the particle decelerates at a rate of g sinβ as it moves from O to P. Suppose that the particle hits P after a time T from the instant of projection. During this time the particle moves up from O to P along the incline with a deceleration g sinβ and moves to and fro perpendicular to the incline.

Considering the motion along Y-axis, the displacement y of the particle during time t (= T) perpendicular to the plane is zero.

### How to Calculate Time of Flight ?

$ \displaystyle y = v_0 sin(\theta – \beta) t – \frac{1}{2} (g cos\beta) t^2 $

and putting y = 0 when t = T we obtain,

$ \displaystyle T = \frac{2 v_0 sin(\theta – \beta)}{g cos\beta}$

### How to Calculate Horizontal Range ?

Now considering the motion along X-axis:

$ \displaystyle x = v_0 cos(\theta – \beta) t – \frac{1}{2} (g sin\beta) t^2 $

Putting , x = R for t = T and substituting the obtained value of T we write

$ \displaystyle R = v_0 cos(\theta – \beta)(\frac{2 v_0 sin(\theta – \beta)}{g cos\beta}) – \frac{1}{2} (g sin\beta) (\frac{2 v_0 sin(\theta – \beta)}{g cos\beta})^2 $

On simplifying ;

$ \displaystyle R = \frac{v_0^2 [sin(2\theta – \beta)-sin\beta] }{g cos^2 \beta} $

### Maximum Range :

Range R is maximum when sin(2θ – β) is maximum, that is equal to one:

$ \displaystyle R_{max} = \frac{v_0^2 [1 -sin\beta] }{g cos^2 \beta} $

$ \displaystyle R_{max} = \frac{v_0^2 [1 -sin\beta] }{g (1 – sin\beta)(1 + sin\beta)} $

$ \displaystyle R_{max} = \frac{v_0^2 }{g (1 + sin\beta)} $

Similarly when the particle is projected down the plane the corresponding range is given as

$ \displaystyle R_{max} = \frac{v_0^2 }{g (1 – sin\beta)} $

Finding the angle θ for maximum range when projected up and down the plane, for

θ = (π/4 + β/2), (π/4 – β/2) it can be found that

$\displaystyle \frac{1}{R_{max}} + \frac{1}{R’_{max}} = \frac{1}{R } $

Where R = maximum range of the projectile on the horizontal plane for same speed of projection.