Helicopter M climbs at 420 km/h along a path making 60 degrees with the horizontal. After 0.5 minutes, is M higher or lower than helicopter N, which is at an altitude of 4.5 km?

Step-by-step solution

1. 1

   Resolve the speed into its vertical component. Only motion perpendicular to the ground adds altitude, and that component is the speed times the sine of the elevation angle:

   $v_y=420\sin 60^{\circ}=420\cdot\frac{\sqrt3}{2}=210\sqrt3\ \text{km/h}$

   Numerically $210\sqrt3=363.731$ km/h.

2. 2

   Convert half a minute into hours. The speed is in km per hour, so the time must be too:

   $0.5\ \text{min}=\frac{0.5}{60}=\frac{1}{120}\ \text{h}$

   Half a minute is a very short time, which is why the answer will turn out small.

3. 3

   Multiply speed by time.

   $h_M=210\sqrt3\cdot\frac{1}{120}=\frac{210}{120}\sqrt3=1.75\sqrt3\ \text{km}$

4. 4

   Evaluate the surd.

   $1.75\sqrt3=1.75\times 1.7320508=3.0311\ \text{km}$

5. 5

   Compare with the other helicopter. $h_N=4.5$ km, and

   $3.0311<4.5$

   so $M$ is lower than $N$ - the opposite conclusion to the faster, longer-flying case, purely because of the shorter time.

6. 6

   Cross-check along the path. In $\tfrac1{120}$ h at $420$ km/h the helicopter travels $3.5$ km along its path; the vertical share is $3.5\sin60^{\circ}=3.5(0.86603)=3.0311$ km ✓.