Projectile Motion — Question 3

Question 3

The apex is not a zero-acceleration point

A projectile leaves the origin with $\vec v_0=(12,16)$ m/s. 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 time, position and speed at the apex.

2. Find the acceleration there and explain what changes immediately before and after the apex.

3. Would a strictly vertical launch have zero acceleration at its apex? Distinguish the velocity condition.

Question 3 - Solution

The apex is not a zero-acceleration point

(a) Set the vertical velocity to zero.

$$0=v_y=16-10t_a\ \Rightarrow\ t_a=1.6\,\mathrm s,\quad\boxed{(x_a,y_a)=(19.2,12.8)\,\mathrm m,\ v_a=12\,\mathrm{m/s}}$$ $v_y=(16-10t/\mathrm s)\ \mathrm{m/s}$ vanishes at 1.6 s. Then $x=19.2$ m and $y=16(1.6)-5(1.6)^2=12.8$ m. Velocity is $(12,0)$ m/s, so apex speed is 12 m/s.

(b) Retain gravity at every point.

$$a_y=\frac{dv_y}{dt}=-10\,\mathrm{m/s^2},\quad v_y(t_a-\epsilon)=10\epsilon,\quad v_y(t_a+\epsilon)=-10\epsilon$$ Acceleration remains $(0,-10)$ m/s$^2$. Just before the apex $v_y>0$; just after it $v_y<0$. The y velocity changes continuously through zero because its derivative stays negative. No pause or switch in gravity is required.

(c) Zero velocity does not eliminate its derivative.

$$v_{0x}=0:\quad\vec v(t_a)=\vec0,\quad\vec a(t_a)=-g\hat j\ne\vec0$$ For a vertical launch $v_x=0$, so the entire velocity is momentarily zero at the apex. Acceleration is still downward with magnitude g. A horizontal component distinguishes a moving apex from a momentarily stationary one; it does not change gravity.