Craft M climbs at 600 km/h along a path making 60 degrees with the horizontal. After 1.8 minutes, is M higher or lower than craft N, which is at an altitude of 10 km?

Step-by-step solution

1. 1

   Take only the vertical part of the velocity. The $600$ km/h is speed along the slanted path, not upward speed. Height depends on the vertical component, which in a right triangle with the path as hypotenuse is the side opposite the $60^{\circ}$ angle:

   $v_y=600\sin 60^{\circ}$

   Using $600$ directly would overstate the altitude by about $15\%$.

2. 2

   Evaluate the sine exactly.

   $\sin 60^{\circ}=\frac{\sqrt3}{2}\ \Longrightarrow\ v_y=600\cdot\frac{\sqrt3}{2}=300\sqrt3\ \text{km/h}$

   Keeping $\sqrt3$ exact avoids rounding drift; numerically $300\sqrt3=519.615$ km/h.

3. 3

   Convert the time to match the speed units. The speed is per hour but the time is in minutes:

   $1.8\ \text{min}=\frac{1.8}{60}=0.03\ \text{h}$

4. 4

   Multiply to get the height climbed.

   $h_M=v_y t=300\sqrt3\times 0.03=9\sqrt3\ \text{km}\approx 15.588\ \text{km}$

5. 5

   Compare with the other craft. $h_N=10$ km, and

   $15.588>10$

   so $M$ is higher.

6. 6

   Check the estimate a second way. In $0.03$ h at $600$ km/h the craft covers $18$ km along the path; the vertical share is $18\sin60^{\circ}=18(0.8660)=15.588$ km ✓ - the same figure by a different route, and comfortably above $10$ km.