Gravity helps at the tiny airport

Classical Mechanics Level 1

Lukla airport in Nepal is one of the strangest in the world. Built to support tourism to the Himalayas, the airport has a single landing runway. What is more, the runway is only 20 m wide, 450 m long, and a 2,800 m cliff at the runway's end, leaving little room for error. In fact, the airport can only be used by so-called Short Takeoff and Landing planes (STOL). Helping somewhat is a 12% incline in the runway from start to finish, so that planes rise through over the course of their deceleration.

Suppose a STOL plane's landing speed is 45 m/s ( $\approx$ 100 mph). Neglecting any other effects like wind flaps, or drag, how small will the plane's velocity (in m/s) be at the top of the runway?

Details

The runway itself is 450 m long, i.e. if you walked from the bottom to the top, you'd walk 450 m along the runway.
An $f$ % incline means that if you walk a distance $d$ along an incline, your rise is given by $fd/100$ .

The answer is 31.0728.

1 solution

Pranshu Gaba
Apr 30, 2015

We can use the work-kinetic energy theorem .

$W_\textrm{net} = \frac12mv_f^2-\frac12mv_i^2$

Only gravity is doing work here, so $W_{\textrm{net}} = - mg \Delta h$ . Hence,

$- mg \Delta h = \frac12mv_f^2-\frac12mv_i^2$

$- g \Delta h = \frac12v_f^2-\frac12v_i^2$

$v_f = \sqrt{v_i^2 - 2g \Delta h}$

Now we put in the values:

$g = 9.8 \mathrm{m s}^{-2}, ~~~ \Delta h = 450 \times \sin \theta = 450 \times 0.12 \mathrm{ m}$

$v_i = 45 \mathrm{m s}^{-1}$ .

After substituting these values, we get

$\boxed{v_i \approx 31.07 \mathrm{ m s}^{-1}} ~~~ _\square$

I make it 31.09?

ben handley - 6 years, 1 month ago

Can be approximated to 30

Syomantak Chaudhuri - 5 years, 4 months ago

Don't you have to change g? Like g times sen (angle)?

Martin Kolton - 5 years, 2 months ago

Would help to be clear that you used g = 9.8 m/s².

Ed Sirett - 4 years, 6 months ago

Gravity helps at the tiny airport

The answer is 31.0728.

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