Projectiles and variable-acceleration motion

What you’ll learn

• Resolve projectile motion into horizontal and vertical parts.
• Differentiate and integrate position and velocity vectors.

Understand the idea

Without air resistance, a projectile has constant horizontal velocity and constant downward acceleration $g$. Resolve its initial velocity using the launch angle, then use the same time in both directions. The usual range and flight-time shortcuts assume launch and landing are at the same height.

When position or velocity is a function of time, differentiate each vector component to move from position to velocity to acceleration. Integrate in the reverse direction and use the given initial conditions to find every constant. Speed is the magnitude of velocity, and maximum height occurs when vertical velocity is zero.

Key formulae

$$x=u\cos\theta\,t,\qquad y=u\sin\theta\,t-\frac12gt^2$$

$$v_x=u\cos\theta,\qquad v_y=u\sin\theta-gt$$

$$\mathbf v=\frac{d\mathbf r}{dt},\qquad\mathbf a=\frac{d\mathbf v}{dt}$$

$$R=\frac{u^2\sin2\theta}{g}\quad\text{(equal launch and landing heights)}$$

Worked example

A projectile has initial velocity $(12\mathbf i+14.7\mathbf j)\,\mathrm{m\,s^{-1}}$ from ground level. Find its horizontal range on returning to ground level. Take $g=9.8\,\mathrm{m\,s^{-2}}$.

1. Vertical displacement is $y=14.7t-4.9t^2$.
2. Set $y=0$ and take the positive root $t=3\,\mathrm s$.
3. Horizontal displacement is $12t=36\,\mathrm m$.

Answer: $36\,\mathrm m$.