# How far along an inclined plane does the projectile land?

A particle is projected with an initial speed of $35 m/s$ from the base of a plane inclined at an angle $\varphi$ to the horizontal where $\varphi=\arctan(3/4)$. If the projectile is initially angled at $\theta$ to the horizontal, where $\tan \theta =5/2$ , ﬁnd how far along the plane the projectile lands, and the time of ﬂight

ANS: 75.431 m, approx. 4.64 s

I can get the position vector for the particle but that is as far as i can get on my own. I don't know how to solve for the point of collision with the inclined plane and everything i try doesn't match the answer in the book.

## 1 Answer

Hint. The particle is initially in the origin. The inclined plane is represented by the line $y=\frac{3x}{4}$. The trajectory of the particle is given by the parametric curve: $$x(t)=v\cos(\theta)t,\quad y(t)=v\sin(\theta)t-\frac{gt^2}{2}$$ or in the implicit form $$y=\tan(\theta)x-\frac{gx^2(1+\tan^2(\theta))}{2v^2}= \frac{5x}{2}-\frac{29 g x^2}{9800}$$ where $v=35 m/s$, $\tan(\theta)=5/2$ and $g=9.8 m/s^2$.

Are you able to solve the problem now?

P.S. After solving the system given by the parabola and the inclined line, you should find that the impact point is approximately $(60.34, 45.26)$ whose distance from the origin is about $75.43 m$.