Not a blog
Projectile Motion
In a projectile motion of an object imparted a velocity (v), the key assumptions are:
- no air resistance to the motion of the object
- earth is flat
- acceleration due to gravity is uniform everywhere
Motion of the object along x and y direction are independent of each other. The important quantities in a projectile motion are R = Horizontal Range, H = Maximum Height above ground, and T = Time of flight.
For a particle projected with velocity ‘v’ and at an angle ‘θ’ with the horizontal.
Here’s the full derivation using the intersection method — a beautifully elegant approach where we treat the projectile as a parabola and the incline as a line, then find where they meet.
Setup
Set up coordinates at the launch point. The projectile is launched with speed $v_0$ at angle $\alpha$ above the inclined plane, which itself makes angle $\beta$ with the horizontal. So the absolute launch angle with the horizontal is $(\alpha + \beta)$.
Equations of motion (standard x–y axes, horizontal and vertical):
\(x = v_0 \cos(\alpha+\beta)\, t\) \(y = v_0 \sin(\alpha+\beta)\, t - \tfrac{1}{2}g t^2\)
Eliminating $t$ from these gives the parabola:
\[y = x\tan(\alpha+\beta) - \frac{g x^2}{2v_0^2\cos^2(\alpha+\beta)}\]The incline as a line
The inclined plane passing through the origin at angle $\beta$:
\[y = x\tan\beta\]Finding the intersection
Set the parabola equal to the line:
\[x\tan\beta = x\tan(\alpha+\beta) - \frac{gx^2}{2v_0^2\cos^2(\alpha+\beta)}\]One solution is $x = 0$ (the launch point). For the other intersection (landing point), divide through by $x$:
\[\tan\beta = \tan(\alpha+\beta) - \frac{gx}{2v_0^2\cos^2(\alpha+\beta)}\]Solving for $x$:
\[x = \frac{2v_0^2\cos^2(\alpha+\beta)}{g}\bigl[\tan(\alpha+\beta) - \tan\beta\bigr]\]Using the identity $\tan A - \tan B = \dfrac{\sin(A-B)}{\cos A \cos B}$ with $A=\alpha+\beta$, $B=\beta$:
\[x = \frac{2v_0^2\cos^2(\alpha+\beta)}{g} \cdot \frac{\sin\alpha}{\cos(\alpha+\beta)\cos\beta} = \frac{2v_0^2\cos(\alpha+\beta)\sin\alpha}{g\cos\beta}\]Converting to range along the slope
The range $R$ along the incline relates to the horizontal landing distance by $x = R\cos\beta$, so $R = x/\cos\beta$:
\[\boxed{R = \frac{2v_0^2 \sin\alpha\,\cos(\alpha+\beta)}{g\cos^2\beta}}\]Using the product-to-sum identity $2\sin A\cos B = \sin(A+B)+\sin(A-B)$:
\[R = \frac{v_0^2}{g\cos^2\beta}\bigl[\sin(2\alpha+\beta) - \sin\beta\bigr]\]Maximum range
$R$ is maximised when $\sin(2\alpha+\beta)=1$, i.e. $2\alpha+\beta = 90°$, giving the optimal launch angle:
\[\alpha_\text{opt} = \frac{90°-\beta}{2} = 45° - \frac{\beta}{2}\]And the maximum range on the incline:
\[R_\text{max} = \frac{v_0^2}{g\cos^2\beta}(1-\sin\beta) = \frac{v_0^2}{g(1+\sin\beta)}\]Let me illustrate the geometry of this intersection:—
Summary of the method
The key insight is elegant: instead of computing the time of flight and range separately, we treat the problem purely geometrically — the projectile traces a parabola, the incline is a straight line through the origin, and the range is simply the distance from the origin to their non-trivial intersection.
This gives us $R = \dfrac{2v_0^2 \sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta}$, which reduces correctly to the flat-ground formula $R = \dfrac{v_0^2\sin 2\alpha}{g}$ when $\beta = 0$.