Projectile Motion: the Angle for Maximum Range

The question

What launch angle throws the furthest?

45°, when the ball lands at the same height it was thrown from. The range is $R={\displaystyle \frac{2v^2\sin \alpha \cos \alpha }{g}}$, and $\sin \alpha \cos \alpha$ is largest at 45°, where the horizontal and vertical parts of the velocity are equal. Angles the same distance either side of 45°, such as 30° and 60°, land in the same place.

In words, then in symbols

• The horizontal part is the whole velocity times the cosine of the angle.

  $v_x=v\cos \alpha$

• The vertical part is the whole velocity times the sine of the angle.

  $v_y=v\sin \alpha$

• Acceleration is how fast velocity changes: the change in velocity divided by the time it takes.

  $a={\displaystyle \frac{v-u}{t}}$

• The range is twice the speed squared, times the sine and the cosine of the launch angle, divided by $g$.

  $R={\displaystyle \frac{2v^2\sin \alpha \cos \alpha }{g}}$

How to solve a projectile question

1. Draw the launch and mark the angle.
2. Split the launch velocity: $v_x=v\cos \alpha$ and $v_y=v\sin \alpha$.
3. Gravity is the only force: vertically the acceleration is $-g$, and horizontally $v_x$ never changes.
4. In the vertical direction, find the time to the top from $v_y=0$, then double it.
5. In the horizontal direction, range = $v_x$ × total time.