Projectile Motion — Question 10

Question 10

Test the ideal model with equal-time position data

At 0, 1 and 2 s, positions are $(0,0)$, $(6,8)$ and $(12,6)$ m. Assume constant acceleration between samples. Use fixed horizontal x and upward y, neglect air resistance, and take uniform $g=10\ \mathrm{m/s^2}$. After release, $\vec a=-g\hat\jmath$. A second data set changes only the last x coordinate to 11 m.

1. Infer acceleration and launch velocity for the original data.

2. Check the original data against the ideal projectile assumptions.

3. Infer the changed horizontal acceleration for the second data set and explain why it cannot be explained merely by a different constant horizontal launch speed.

Question 10 - Solution

Test the ideal model with equal-time position data

(a) Use equal-interval second differences.

$$\vec a=\frac{\vec r_2-2\vec r_1+\vec r_0}{h^2}=(0,-10)\,\mathrm{m/s^2},\quad\boxed{\vec v_0=\frac{\vec r_1-\vec r_0}{h}-\tfrac12\vec ah=(6,13)\,\mathrm{m/s}}$$ For interval $h=1$ s, $\vec a=[\vec r(2)-2\vec r(1)+\vec r(0)]/h^2=(0,-10)$ m/s$^2$. From $\vec r(1)-\vec r(0)=\vec v_0h+\tfrac12\vec a h^2$, the launch velocity is $(6,13)$ m/s.

(b) Compare the inferred vector with gravity.

$$\Delta x_1=6\,\mathrm m=\Delta x_2,\quad\Delta y_2-\Delta y_1=-2-8=-10\,\mathrm m=-gh^2$$ The acceleration matches $(0,-g)$ and the horizontal step is 6 m in each second. The three samples are consistent with the assumed constant-acceleration ideal projectile model. They alone cannot prove that no smaller unresolved deviations occurred between samples.

(c) Use the altered horizontal second difference.

$$a_x'=\frac{11-2(6)+0}{1^2}=-1\,\mathrm{m/s^2}\ne0$$ $a_x=(11-2\cdot6+0)\ \mathrm m/(1\ \mathrm s)^2=-1$ m/s$^2$. A different constant launch speed would still give equal horizontal increments, but these increments are 6 and 5 m. Under the constant-acceleration assumption, the second set needs a horizontal acceleration or a measurement correction.