Question

An aircraft is climbing at a 30 degree angle to the horizontal. How fast is the aircraft gaining altitude if it's speed is 500 miles per hour?

Answer 1

Answer:

The aircraft is gaining altitude at $250\frac{mi}{h}$

Explanation:

Draw a vector diagram. (Right angle triangle speed as the hypotenuse)

Let

x -----> horizontal component of the speed

y ----> vertical component of the speed

s ----> speed in miles per hour

we know that

$x=cos(30\°)(s)$

$y=sin(30\°)(s)$

we have

$s=500\ \frac{mi}{h}$

$cos(30\°)=\frac{\sqrt{3}}{2}$

$sin(30\°)=\frac{1}{2}$

substitute

$x=\frac{\sqrt{3}}{2}(500)$

$x=250\sqrt{3}\frac{mi}{h}$

$y=\frac{1}{2}(500)$

$y=250\frac{mi}{h}$

therefore

The aircraft is gaining altitude at $250\frac{mi}{h}$

Answer 2

The speed rate at which the aircraft is gaining altitude is;

dx/dt = 250 miles per hour

Since the aircraft is climbing at 30° to the horizontal. it means there will be a vertical component which will be the altitude and also a horizontal component which will be the horizontal distance covered. while the hypotenuse will be how far it has climbed.

Let us call the vertical component x and the hypotenuse component z.

Thus, from trigonometric ratios, we can say that;

sin 30 = x/z

We are looking for how fast the aircraft is gaining altitude. Thus, using Liebniz notation, we can say this rate is; dx/dt. Thus;

sin 30 = (dx/dt)/(dz/dt)

0.5 = (dx/dt)/(dz/dt)

0.5 × (dz/dt) = dx/dt

Its' speed is 500 mph. This means dz/dt = 500

Thus;

dx/dt = 0.5 × 500

dx/dt = 250 mph

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