Projectile Motion — Question 8

Question 8

Gravity makes equal downward departures from a tangent line

A projectile is launched with $\vec v_0=(8,6)$ m/s from the origin. Define the reference point $\vec r_{\mathrm{straight}}(t)=\vec v_0t$, which would occur if acceleration vanished. 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. Find the vector difference between actual and reference positions at time t.

2. Evaluate the two positions and their separation at 1 s.

3. Does the downward separation depend on horizontal launch speed? Explain why the reference line is not a second real path under the stated gravity.

Question 8 - Solution

Gravity makes equal downward departures from a tangent line

(a) Subtract the two vector histories.

$$\boxed{\vec r-\vec r_{\mathrm{straight}}=\left(\vec v_0t-\tfrac12gt^2\hat j\right)-\vec v_0t=-\tfrac12gt^2\hat j}$$ Actual position is $\vec r=\vec v_0t-\tfrac12gt^2\hat\jmath$. Hence $\vec r-\vec r_{\mathrm{straight}}=-\tfrac12gt^2\hat\jmath$: a purely downward departure with no horizontal difference.

(b) Compare at the same time.

$$\vec r_{\mathrm{straight}}(1)=(8,6)\,\mathrm m,\quad\vec r(1)=(8,1)\,\mathrm m,\quad\Delta\vec r=(0,-5)\,\mathrm m$$ At 1 s the reference point is $(8,6)$ m and the actual projectile is $(8,1)$ m. Their separation is 5 m downward. This compares simultaneous positions, not the shortest distance from the parabola to the reference line.

(c) Identify the counterfactual assumption.

$$\frac{\partial}{\partial v_{0x}}\left(-\tfrac12gt^2\right)=0,\quad\vec a_{\mathrm{straight}}=\vec0\ne-g\hat j$$ The departure depends only on g and elapsed time for the same initial position and velocity. Horizontal launch speed changes x but not the vertical shortfall at a fixed time. The straight reference path omits gravity and therefore serves as a construction, not another possible trajectory under the full stated model.