Aircraft Q climbs at 450 km/h along a path making an angle of 45° with the horizontal. After 0.7 minutes, is Q higher or lower than aircraft P, which is at an altitude of 5 km?

Step-by-step solution

1. 1

   Split the velocity into horizontal and vertical parts. The speed $450$ km/h is measured along the flight path, not straight up. In the right triangle formed by the path, the ground and the altitude, the altitude is the side opposite the climb angle, so

   $v_y=450\sin 45^{\circ}.$

   Using the speed itself as the height rate is the classic error here.

2. 2

   Evaluate the vertical speed. With $\sin 45^{\circ}=\tfrac{\sqrt2}{2}$,

   $v_y=450\cdot \tfrac{\sqrt2}{2}=318.1981\ \text{km/h}.$

3. 3

   Convert the time to hours. The speed is per hour but the time is in minutes, so

   $t=\frac{0.7}{60}=0.0116667\ \text{h}.$

   Mixing units at this point is the second classic error.

4. 4

   Compute the altitude gained.

   $h_{Q}=v_y\,t=318.198\times 0.0116667=3.71231\ \text{km}.$

   Equivalently, fly $5.25$ km along the path and take the fraction $\sin 45^{\circ}$ of it. Here $\sin 45^{\circ}=\tfrac{\sqrt2}{2}$ is irrational, so the exact height is $5.25\cdot\tfrac{\sqrt2}{2}=\tfrac{21\sqrt2}{8}$ km, numerically $3.712311$ km.

5. 5

   Compare with aircraft $P$. Since $3.7123 < 5$, aircraft $Q$ is lower than aircraft $P$. To overtake $P$ it would need $t>0.9428$ minutes at the same speed and angle.