An aircraft climbs at 500 km/h along a straight path that makes an angle of 30 degrees with the horizontal. After 1.2 minutes, is it higher or lower than a second aircraft flying at an altitude of 7 km?

Step-by-step solution

1. 1

   Separate the speed into components. The $500$ km/h is measured along the flight path, not straight up, so it cannot be multiplied by the time directly. Drawing the right triangle whose hypotenuse is the flight path, the vertical leg is opposite the $30^{\circ}$ angle, so the vertical (climb) speed is

   $v_{y}=v\sin\theta=500\sin 30^{\circ}.$

   Use sine, not cosine: sine pairs the angle with the opposite side, which is the height.

2. 2

   Evaluate the vertical speed. Since $\sin 30^{\circ}=\tfrac12$,

   $v_{y}=500\cdot\frac{1}{2}=250\ \text{km/h}.$

   Only half the aircraft's speed is being spent gaining altitude; the other component, $500\cos 30^{\circ}\approx 433$ km/h, moves it horizontally.

3. 3

   Match the units of time to the units of speed. The speed is per hour but the time is given in minutes, so convert before multiplying:

   $t=1.2\ \text{min}=\frac{1.2}{60}\ \text{h}=0.02\ \text{h}.$

   Skipping this conversion is the most common error in this type of problem.

4. 4

   Compute the altitude gained.

   $h=v_{y}t=250\times 0.02=5\ \text{km}.$

5. 5

   Compare with the second aircraft. The climbing aircraft is at $5$ km while the other is at $7$ km, and $5<7$, so after $1.2$ minutes the climbing aircraft is still lower. It would need $\dfrac{7}{250}\ \text{h}=1.68$ minutes at this rate to reach $7$ km.