Projectile Motion — Question 5

Question 5

Equal-height velocity symmetry has precise assumptions

A projectile starts with velocity $(9,15)$ m/s at y=0. Use fixed horizontal x and upward y, neglect air resistance, and take uniform $g=10\ \mathrm{m/s^2}$. After release, $\vec a=-g\hat\jmath$. Consider two instants when its height is 10 m.

1. Find the two times and the velocities.

2. Compare speeds and horizontal positions at those times.

3. Explain which conclusion would fail if the two observations were at unequal heights, even with the same ideal model.

Question 5 - Solution

Equal-height velocity symmetry has precise assumptions

(a) Solve the vertical quadratic and retain both roots.

$$5t^2-15t+10=5(t-1)(t-2)=0\ \Rightarrow\ \boxed{t_1=1\,\mathrm s,\quad t_2=2\,\mathrm s}$$ $10=15t-5t^2$ gives $t=1$ s and $t=2$ s. The velocities are $(9,5)$ and $(9,-5)$ m/s. The first event is ascending and the second descending.

(b) Equal speed does not imply equal vector or position.

$$\vec v_1=(9,5),\quad\vec v_2=(9,-5)\quad\mathrm{m/s},\quad v_1=v_2=\sqrt{106}\,\mathrm{m/s},\quad x_1=9,\ x_2=18\,\mathrm m$$ Both speeds are $\sqrt{81+25}=\sqrt{106}\approx10.296$ m/s, but the vertical directions are opposite. Horizontal positions are 9 and 18 m. The two states have equal height and speed, not identical velocity or location.

(c) Use the vertical speed-height relation.

$$v^2=v_x^2+v_{0y}^2-2g(y-y_0)\ \Rightarrow\ v_2^2-v_1^2=-2g(y_2-y_1)$$ $v_y^2=v_{0y}^2-2g(y-y_0)$ fixes the vertical speed magnitude at a given height. Unequal heights generally give unequal $|v_y|$ and total speeds. The horizontal velocity remains the same at both observations regardless of height, provided the ideal assumptions hold.