Projectile Motion — Question 6

Question 6

Horizontal speed changes the trajectory, not the vertical clock

Projectiles A and B leave the same point simultaneously with velocities $(5,12)$ and $(15,12)$ m/s. They land on the launch-height plane. 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 their common apex time and return time.

2. Find both ranges and maximum heights.

3. At any common airborne time, find B’s position relative to A. Can they meet again after release?

Question 6 - Solution

Horizontal speed changes the trajectory, not the vertical clock

(a) Use their shared vertical equation.

$$t_a=\frac{12}{10}=1.2\,\mathrm s,\quad\boxed{T=\frac{2(12)}{10}=2.4\,\mathrm s}$$ Both satisfy $y=12t-5t^2$ with numerical t in seconds. The apex time is 1.2 s and the positive return time is 2.4 s. Their vertical histories are identical.

(b) Let horizontal speed set the spatial spread.

$$R_A=5(2.4)=12\,\mathrm m,\quad R_B=15(2.4)=36\,\mathrm m,\quad H=\frac{12^2}{20}=7.2\,\mathrm m$$ Ranges are $5(2.4)=12$ m and $15(2.4)=36$ m. Both maximum heights are $12^2/(2g)=7.2$ m. Tripling horizontal speed triples the range while leaving this vertical motion unchanged.

(c) Subtract simultaneous position vectors.

$$\vec r_{B/A}=(15-5)t\hat i=(10\,\mathrm{m/s})t\hat i=\vec0\ \Leftrightarrow\ t=0$$ $\vec r_B-\vec r_A=(10\ \mathrm{m/s})t\hat\imath$. The vertical difference is zero, but horizontal separation is positive for every $t>0$. They meet only at release within the stated flight interval. Equal height at the same time is not a collision condition by itself.