Hello demode Originally Posted by

**demode** Hello Grandad,

I understood your methd.

But we know the glide angle:

$\displaystyle (mg)sin \theta = drag force$

$\displaystyle \theta = sin^{-1} \frac{1300}{1300 \times 9.81} = 5.85$

So now that we know the glide angle is 5.85°, how do we find the rate of change of y?

Having just re-read my previous answer, I realise I didn't word it very well. There are, of course, three forces acting on the aircraft, the third one being the lift force. But this is at right-angles to the direction of motion, and hence does no work. Therefore my answer is correct. PE is being lost at the rate of 47306 J/s.

To get the rate of change of the vertical height $\displaystyle y$, simply divide this rate of change of PE by $\displaystyle mg$. (Can you see why?) So the rate of change of $\displaystyle y$ is $\displaystyle -\frac{47306}{1300\times9.81}=-3.71$ m/s

The equation you have written down is correct, as is the angle of the flight path. This gives an alternative way of calculating the rate of change of $\displaystyle y$. It is simply the vertical component of the speed, in m/s. Which is:$\displaystyle -\frac{131 \times 1000}{3600}\sin5.85^o = -3.71$ m/s

I haven't seen the original wording of the question, but if (as you suggest) it asks for the rate of loss of PE and then the rate of change of $\displaystyle y$, I think they expect the first method above to be used. But the second is equally valid.

Grandad