Rates of Change question (1 Viewer)

[Chris100]

The equation of the path of a bullet fired into the air is y=-20x(x-20), where x and y are displacements in metres horizontally and vertically from the origin. The bullet is moving horizontally at a constant rate of 1/2 ms^-1.
How high does the bullet go, and how far away does it land?

Please have a go at the maximum x-displacement

 

[braintic]

The equation of the path of a bullet fired into the air is y=-20x(x-20), where x and y are displacements in metres horizontally and vertically from the origin. The bullet is moving horizontally at a constant rate of 1/2 ms^-1.
How high does the bullet go, and how far away does it land?

Please have a go at the maximum x-displacement

The roots are x=0 and x=20. So it starts at ground level at x=0, and finishes at ground level (ie. lands) at x=20.

 

[anomalousdecay]

The roots are x=0 and x=20. So it starts at ground level at x=0, and finishes at ground level (ie. lands) at x=20.

If OP is still stuck, highlight below:

The maximum height occurs at x=10m, since the parabola given is symmetrical.

Also, instead you can just derive and find the maximum point (you will get x=10m).

now substitute x=10m into the equation and you will get that the maximum height is:

y= -20(10m)(-10m)

y= 2000m

This question is very odd though. The answer I got is basically like saying the bullet was fired up straight in the air.

 

[integral95]

and the maximum height is the maximum value of the equation of the parabola i.e at the centre of the parabola (so you should know what x value to sub)

 

[anomalousdecay]

and the maximum height is the maximum value of the equation of the parabola i.e at the centre of the parabola (so you should know what x value to sub)

As I said above.

 

[braintic]

As I said above.

I'm afraid you didn't say anything above. There is only white space between your first and last lines.

 

[funnytomato]

I'm afraid you didn't say anything above. There is only white space between your first and last lines.

 

[braintic]

OK, interesting. It wasn't obvious to me what that meant.

 

[anomalousdecay]

OK, interesting. It wasn't obvious to me what that meant.

I thought you already knew. Its accustomed to allow our hints and clues help them first, and if they still have problems, they can highlight and check below, or check their answer with ours.

And thanks funnytomato for doing the honours of showing others how to find the answer.