Moving object problems

So far the observer and the object have both been at rest. The next family of problems lets one of them move , a plane crossing the sky, a boat sailing along a river, a car approaching a tower. Now we use the same right-triangle ideas, but extract a distance covered and divide by a time taken to get a speed (or vice versa). This is just kinematics glued onto trigonometry.

Movement problems are slightly more demanding than static ones for a single reason: you must hold two configurations of the diagram in your head , the moment when the first angle is observed, and the moment when the second is observed , and use both to set up equations. A neat trick is to draw both positions on the same diagram, distinguishing them by labels (Position 1, Position 2) or colours.

Generic template

Let a moving object be observed at time $t = 0$ at one position and at time $t = T$ at another. Let the elevations (or depressions) at the two moments be $\alpha$ and $\beta$ , and let the height of the path (assumed constant) be $H$ .

If we know $H$ , we compute horizontal distances at both moments:

$x_1 = \frac{H}{\tan\alpha}, \qquad x_2 = \frac{H}{\tan\beta}.$

The distance covered is $|x_1 - x_2|$ . The speed is $|x_1 - x_2|/T$ .

Aeroplane crossing overhead

If a plane flies horizontally at constant height $H$ and its elevation from an observer is $\alpha$ at $t = 0$ and $\beta$ at $t = T$ ,

$\text{speed} = \frac{H(\cot\alpha - \cot\beta)}{T} \quad (\text{if moving away}).$

(If moving toward, swap $\alpha$ and $\beta$ so the answer is positive.)

For $\alpha = 60^\circ, \beta = 30^\circ$ : $\cot 60^\circ = 1/\sqrt 3$ , $\cot 30^\circ = \sqrt 3$ , difference $= \sqrt 3 - 1/\sqrt 3 = 2/\sqrt 3 = 2\sqrt 3/3$ . So speed $= 2\sqrt 3 H / (3T)$ .

Boats and cars

A car (or boat) on the ground, observed from the top of a tower of height $H$ . Let the depression at $t = 0$ be $\alpha$ and at $t = T$ be $\beta$ , with the object approaching the tower so $\alpha < \beta$ . Then

$x_1 = H/\tan\alpha, \quad x_2 = H/\tan\beta, \quad \text{speed} = \frac{x_1 - x_2}{T}.$

If the speed and one observation are known, you can find the time it takes to reach the foot of the tower (set $x_2 = 0$ , so $\beta = 90^\circ$ ).

Worked examples

Example 1. A plane flies horizontally at $3000$ m. Its elevation from a point on the ground changes from $60^\circ$ to $30^\circ$ in $15$ seconds. Find the speed of the plane.

$x_1 = 3000/\tan 60^\circ = 3000/\sqrt 3 = 1000\sqrt 3$ . $x_2 = 3000/\tan 30^\circ = 3000\sqrt 3$ .

Distance covered $= 3000\sqrt 3 - 1000\sqrt 3 = 2000\sqrt 3$ m.

Speed $= 2000\sqrt 3 / 15 = 400\sqrt 3/3 \approx 230.94$ m/s.

In km/h: $\approx 831.4$ km/h.

Example 2. A man on top of a $100$ -m cliff sees a car. The angle of depression changes from $30^\circ$ to $60^\circ$ as the car moves toward him. How far did the car travel?

$x_1 = 100/\tan 30^\circ = 100\sqrt 3$ . $x_2 = 100/\tan 60^\circ = 100/\sqrt 3 = 100\sqrt 3/3$ .

Distance covered $= 100\sqrt 3 - 100\sqrt 3/3 = 200\sqrt 3/3 \approx 115.5$ m.

Example 3. From the top of a $75$ -m lighthouse, the depressions of two ships approaching it in the same straight line are $30^\circ$ and $45^\circ$ . After how much time will the farther ship reach the foot of the lighthouse, given the nearer ship reached in $5$ minutes?

Initial distances: $x_1 = 75/\tan 30^\circ = 75\sqrt 3$ (farther), $x_2 = 75/\tan 45^\circ = 75$ (nearer).

The nearer covered $75$ m in $5$ minutes, so speed $= 15$ m/min.

The farther ship has $75\sqrt 3 \approx 129.9$ m to go, at $15$ m/min: time $\approx 8.66$ min.

(Actually re-read the original problem carefully , board questions often assume both ships travel at the same speed; this is the kind of nuance to verify.)

Example 4. A boy flies a kite at a height of $90$ m. He runs in a horizontal direction while letting out string, until the angle the string makes with the horizontal changes from $60^\circ$ to $30^\circ$ . The string is taut throughout. How much extra string did he let out?

Initial string $L_1 = 90/\sin 60^\circ = 90 \cdot 2/\sqrt 3 = 60\sqrt 3$ . Final string $L_2 = 90/\sin 30^\circ = 90 \cdot 2 = 180$ .

Extra string $= 180 - 60\sqrt 3 \approx 180 - 103.9 = 76.1$ m.

Example 5. A man standing on the deck of a ship $10$ m above water-level sees a hill. The angle of elevation of the top of the hill from his eye is $60^\circ$ and the angle of depression of the base of the hill is $30^\circ$ . Find the distance of the hill from the ship and the height of the hill.

Let $d$ = horizontal distance from the man's eye to the hill, $h$ = hill's height above water.

Depression to base: $\tan 30^\circ = 10/d \Rightarrow d = 10\sqrt 3$ m.

Elevation to top: $\tan 60^\circ = (h - 10)/d \Rightarrow h - 10 = 10\sqrt 3 \cdot \sqrt 3 = 30 \Rightarrow h = 40$ m.

So the ship is $10\sqrt 3 \approx 17.3$ m from the hill and the hill is $40$ m tall.

Try it yourself

From a $50$ -m cliff, a boat's depression changes from $45^\circ$ to $60^\circ$ in $2$ s. Find the boat's speed.
The elevation of a plane (height $1500$ m) changes from $30^\circ$ to $60^\circ$ in $20$ s. Find its speed in m/s.
A car approaches a tower of height $40$ m. The depression changes from $30^\circ$ to $45^\circ$ in $10$ s. Find the speed.
A kite is at height $80$ m. The string-to-horizontal angle changes from $30^\circ$ to $45^\circ$ as the boy walks toward the point below the kite. Find the change in string length.
From the top of a $90$ -m hill the angles of depression of two boats on a river (in line, on the same side) are $30^\circ$ and $60^\circ$ . How far apart are the boats?
A balloon rises vertically. Its elevation from a point $100$ m away changes from $30^\circ$ to $60^\circ$ in $4$ minutes. Find the balloon's vertical speed.
From a point $A$ , a car's elevation toward the top of a tower changes from $30^\circ$ to $60^\circ$ as the driver moves $30$ m toward it. Find the tower's height.
A bird flies horizontally at $30$ m. Its elevation from a point on the ground changes from $45^\circ$ to $30^\circ$ in $5$ s. Find its speed.
The angle of elevation of a flying jet, height $h$ , is $60^\circ$ . After $1$ minute it is $30^\circ$ . If the jet flies at $720$ km/h horizontally, find $h$ .
A boat sails away from a $20$ -m cliff. The angles of depression are $45^\circ$ initially and $30^\circ$ after $3$ s. Find the boat's speed.

Pitfalls / Insight

The crucial step is to be honest about whether the object is approaching or moving away; this fixes the sign of the distance and hence the speed. Always state your assumption clearly and check that the larger horizontal distance corresponds to the smaller angle (for elevation/depression of objects on the ground).

Convert units carefully. Board questions occasionally give a height in metres and ask for a speed in km/h , multiply by $18/5$ (or divide by $5/18$ ). Double-check the conversion before circling the final answer.