Projectile Motion Analyzed — Question 8

Question 8

Landing on an incline changes the target equation

A projectile launches from the origin with velocity $(10,20)$ m/s. The ground rises according to $y=x/2$ for x>0, and the projectile lands at its first positive-time intersection with that surface. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. Find the landing time and coordinates.

2. Find the distance along the incline from launch to landing.

3. Check that the trajectory is above the incline between release and landing. Why is the equal-height range formula inapplicable?

Question 8 - Solution

Landing on an incline changes the target equation

(a) Use the surface equation at the same time.

$$20t-5t^2=\frac{10t}{2}\ \Rightarrow\ 5t(3-t)=0\ \Rightarrow\ \boxed{T=3\,\mathrm s,\quad(x,y)=(30,15)\,\mathrm m}$$ $x=10t$ and $y=20t-5t^2$ in SI numerical coordinates. Setting $y=x/2$ gives $20t-5t^2=5t$, or $5t(3-t)=0$. The positive landing time is 3 s, at $(30,15)$ m.

(b) Measure along the actual ground line.

$$s_{\mathrm{incline}}=\sqrt{x^2+y^2}=\sqrt{30^2+15^2}=15\sqrt5\,\mathrm m$$ The surface distance is $\sqrt{30^2+15^2}=15\sqrt5\approx33.541$ m. The horizontal range is 30 m; these are distinct distances with different geometric meanings.

(c) Verify first contact, not just an algebraic intersection.

$$y-\tfrac12x=5t(3-t)>0\quad(0<t<3\,\mathrm s),\qquad y_f=15\,\mathrm m\ne y_0$$ The vertical gap is $y-x/2=5t(3-t)$ m for numerical seconds, positive for $0<t<3$. Thus no earlier contact occurs. Landing is above the launch height, so the equal-height flight time and its derived range formula do not apply.