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.


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 $75.43 m$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.