Range and launch angle: interpreting data

G1.5 G1.4

By the end of this lesson you should be able to

Interpret data relating to the horizontal distance travelled by an object projected at various angles from the horizontal. G1.5
Solve problems involving projectile motion in the absence of drag effects using v_y = u_y + gt, s_y = u_y t + ½gt^2, v_y^2 = u_y^2 + 2gs_y, v_x = u_x and s_x = u_x t. G1.4

Key ideas

For the same launch speed on level ground, range depends on \(\sin 2\theta\): maximum at 45°.
Complementary angles (e.g. 30° and 60°) give the same range but different heights and flight times.
To test the model, linearise: plot range against \(\sin 2\theta\). A straight line through the origin supports the model; the gradient is \(u^2/g\).
Real data falls short at large angles and over long flights because of air resistance and launch-height differences.

Key formulas (QCAA formula book notation)

\[ \text{range} = \frac{u^2 \sin 2\theta}{g} \quad \text{(level ground, no air resistance)} \]

Worked example

A launcher gives a range of 4.0 m at 45°. Predict the range at 15° (same speed, level ground).

Range \(\propto \sin 2\theta\): \(4.0 \times \sin 30^\circ / \sin 90^\circ = 2.0\) m

Common mistakes

PhET: Projectile Data Lab. Collect range data at several launch angles, then look for the angle that gives the maximum range.

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