Projectile Motion — Question 4

Question 4

Recover gravity from a velocity-time record

A projectile has measured velocities $(7,12)$ m/s at 0 s, $(7,2)$ m/s at 1 s and $(7,-8)$ m/s at 2 s. Its initial position is $(0,3)$ m. Use fixed horizontal x and upward y, neglect air resistance, and take uniform $g=10\ \mathrm{m/s^2}$. After release, $\vec a=-g\hat\jmath$.

1. Check whether these records are consistent with ideal projectile motion.

2. Find the position at 2 s and the time of the apex.

3. A further record gives velocity $(6,-18)$ m/s at 3 s. Identify which ideal-model condition it violates.

Question 4 - Solution

Recover gravity from a velocity-time record

(a) Test both component changes.

$$\frac{\vec v(1)-\vec v(0)}1=\frac{\vec v(2)-\vec v(1)}1=(0,-10)\,\mathrm{m/s^2}$$ The horizontal component remains 7 m/s, while the vertical component decreases by 10 m/s each second. The records agree with $a_x=0$, $a_y=-10$ m/s$^2$ and launch velocity $(7,12)$ m/s.

(b) Integrate from the specified release point.

$$\boxed{\vec r(2)=(7(2),3+12(2)-5(2)^2)=(14,7)\,\mathrm m},\quad t_a=12/10=1.2\,\mathrm s$$ $x=7t$ and $y=3+12t-5t^2$ when numerical t is in seconds and coordinates in metres. At 2 s the point is $(14,7)$ m. The apex occurs when $12-10t=0$, giving 1.2 s.

(c) Check the horizontal conservation separately.

$$\vec v_{\mathrm{pred}}(3)=(7,12-30)=(7,-18),\quad\vec v_{\mathrm{obs}}-\vec v_{\mathrm{pred}}=(-1,0)\quad\mathrm{m/s}$$ At 3 s the model predicts $(7,-18)$ m/s. The reported vertical entry agrees, but the horizontal entry is 1 m/s too small. Exact data of that form require an additional horizontal effect or a recording error; they cannot all satisfy the stated ideal model.