Projectile Motion — Question 1

Question 1

Resolve the launch before evolving the motion

A ball is launched from the origin at 20 m/s, $30^\circ$ above horizontal. 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 launch velocity components and the functions x(t), y(t), $v_x(t)$ and $v_y(t)$.

2. Find position and velocity at 1 s.

3. Explain why setting both acceleration components equal to $-g$ would be incorrect.

Question 1 - Solution

Resolve the launch before evolving the motion

(a) Resolve once, then evolve each component.

$$v_{0x}=20\cos30^\circ=10\sqrt3,\quad v_{0y}=20\sin30^\circ=10,\quad x=v_{0x}t,\ y=v_{0y}t-\tfrac12gt^2$$ $v_{0x}=10\sqrt3$ m/s and $v_{0y}=10$ m/s. Thus $x=(10\sqrt3\ \mathrm{m/s})t$, $y=(10\ \mathrm{m/s})t-\tfrac12gt^2$, $v_x=10\sqrt3$ m/s and $v_y=10\ \mathrm{m/s}-gt$.

(b) Use the same time in both equations.

$$\boxed{\vec r(1)=(10\sqrt3,5)\,\mathrm m},\quad\vec v(1)=(10\sqrt3,10-10)=(10\sqrt3,0)\,\mathrm{m/s}$$ At 1 s, position is $(10\sqrt3,5)$ m and velocity is $(10\sqrt3,0)$ m/s. The ball is at its highest point, but still moves horizontally.

(c) Gravity has one specified physical direction.

$$\vec a=\frac{d\vec v}{dt}=(0,-g),\quad\vec a_{\mathrm{incorrect}}=(-g,-g)\ne\vec a$$ The acceleration vector is $(0,-10)$ m/s$^2$. Assigning $a_x=-g$ would invent an additional horizontal acceleration and slow the horizontal motion. Component independence means each equation uses its own acceleration component, not that both components receive the same number.