Height, Range and Time of Flight: MDCAT Physics notes

The three results

For a projectile launched with speed $v_i$ at angle $\theta$ above the horizontal, landing back on the same level (air resistance neglected), three quantities are asked again and again.

Maximum height H

At the top $v_y = 0$. Using $v_y^2 = v_{iy}^2 - 2gH$:

$$H = \frac{v_i^2\sin^2\theta}{2g}$$

Time of flight T

Time to reach the top is $v_i\sin\theta/g$; the fall takes the same time, so

$$T = \frac{2v_i\sin\theta}{g}$$

Example: $v_i = 5\ \text{m s}^{-1}$ at $45^\circ$ gives $T = 2(5)(0.707)/9.8 \approx 0.72\ \text{s}$.

Horizontal range R

Range = horizontal velocity × time of flight $= v_i\cos\theta \times 2v_i\sin\theta/g$. Using $2\sin\theta\cos\theta = \sin2\theta$:

$$R = \frac{v_i^2\sin2\theta}{g}$$

The range is measured horizontally.

Maximum and minimum range

• $R$ is maximum when $\sin2\theta = 1$, i.e. $2\theta = 90^\circ$, so $\theta = 45^\circ$. Then $R_{max} = v_i^2/g$.
• $R$ is zero (minimum) when $\sin2\theta = 0$: at $\theta = 0^\circ$ or $\theta = 90^\circ$ (thrown straight up, it comes straight down).
• Half the maximum range: $\sin2\theta = \tfrac12$, so $2\theta = 30^\circ$ and $\theta = 15^\circ$ (or $75^\circ$).
• Since $R \propto v_i^2$, doubling the speed makes the range four times; $H$ also becomes four times, and $T$ doubles.

Complementary angles give equal range

Because $\sin2\theta = \sin(180^\circ - 2\theta) = \sin 2(90^\circ - \theta)$, angles $\theta$ and $(90^\circ - \theta)$ give the same range at the same speed. Pairs such as $15^\circ$ & $75^\circ$, $30^\circ$ & $60^\circ$, $40^\circ$ & $50^\circ$ and $(45^\circ+\alpha)$ & $(45^\circ-\alpha)$ have equal ranges (ratio 1). The larger angle goes higher and stays in the air longer, but lands at the same point.

For a fixed speed, increasing the angle beyond $45^\circ$ increases the height but decreases the range.

Ratio of height to range

Dividing $H$ by $R$:

$$\frac{H}{R} = \frac{\tan\theta}{4}$$

Key formulas

• $H = \dfrac{v_i^2\sin^2\theta}{2g}$
• $T = \dfrac{2v_i\sin\theta}{g}$
• $R = \dfrac{v_i^2\sin2\theta}{g}$, $\quad R_{max} = \dfrac{v_i^2}{g}$ at $45^\circ$
• $\dfrac{H}{R} = \dfrac{\tan\theta}{4}$; at $45^\circ$, $H_{max\ range} = R_{max}/4$

Common MDCAT traps

• Maximum range needs $\sin2\theta = 1$, not $\sin\theta = 1$.
• Equal ranges come from angles that add to $90^\circ$, not angles that differ by a fixed amount.
• Half of maximum range is at $15^\circ$ (or $75^\circ$), not $22.5^\circ$.
• Range depends on $v_i^2$: doubling speed gives $4R$, not $2R$.
• At $90^\circ$ height is maximum but range is zero.

Quick revision

• Maximum range at $45^\circ$.
• $\theta$ and $90^\circ-\theta$: same range, different heights and times.
• $H/R = \tan\theta/4$.
• Time of flight is twice the time to reach the top.
• Above $45^\circ$: more height, less range.