Projectile Motion — Question 7

Question 7

Eliminate time to reveal a parabolic path

A projectile leaves $(0,5)$ m with velocity $(4,10)$ m/s and is absorbed when it first reaches the floor y=0. Use fixed horizontal x and upward y, neglect air resistance, and take uniform $g=10\ \mathrm{m/s^2}$. After release, $\vec a=-g\hat\jmath$.

1. Eliminate t to find y(x), with dimensions shown.

2. Find the physical horizontal domain before impact.

3. Explain why points with x<0 on the same algebraic parabola are not part of this launched motion.

Question 7 - Solution

Eliminate time to reveal a parabolic path

(a) Solve the horizontal clock for time.

$$t=\frac{x}{4\,\mathrm{m/s}}\ \Rightarrow\ \boxed{y=5\,\mathrm m+\frac52x-\frac{5x^2}{16\,\mathrm m}}$$ $t=x/(4\ \mathrm{m/s})$. Substituting into the vertical equation gives $y=5\ \mathrm m+(5/2)x-[5/(16\ \mathrm m)]x^2$. The quadratic coefficient has inverse-length units so each term is a length.

(b) Stop at the first positive impact time.

$$0=5+10t-5t^2\ \Rightarrow\ T=1+\sqrt2\,\mathrm s,\quad0\le x\le4(1+\sqrt2)\,\mathrm m$$ $0=5+10t-5t^2$ gives $t=1\pm\sqrt2$ s. Only $1+\sqrt2$ s is after launch. Therefore $0\leq x\leq4(1+\sqrt2)$ m, approximately 9.657 m.

(c) A curve equation alone does not encode the launch interval.

$$x=(4\,\mathrm{m/s})t,\quad t\ge0\ \Rightarrow\ x\ge0$$ Since $v_x=4$ m/s and launch occurs at $t=0$, negative x corresponds to negative t under this parametrization. Those points are an extrapolation before release. The floor also truncates the positive-time mathematical curve at impact.