- #1

- 71

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I have:

using pythagoras....

c^2=(1+x-(x^2/40))+x^2

i dont know what to do from there...

thanks for your help

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- Thread starter aricho
- Start date

- #1

- 71

- 0

I have:

using pythagoras....

c^2=(1+x-(x^2/40))+x^2

i dont know what to do from there...

thanks for your help

- #2

Integral

Staff Emeritus

Science Advisor

Gold Member

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Since h is the height of the ball why not try to solve your equation for x when h=0?

- #3

HallsofIvy

Science Advisor

Homework Helper

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Why Pythagoras? Do you have a right triangle here? Do you even have straight lines here?

- #4

BobG

Science Advisor

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- #5

VietDao29

Homework Helper

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I just want to make it a little bit clearer...

this means that you can found the height of the cricket ball by plugging the x - the horizontal distance the ball have travelled into h = 1 + x - (xaricho said:" the height "h" meters of a cricket ball after being struck by a batsman is given by the equation h=1+x-(x^2/40) where x meters is the horisontal distance travelled by the ball from the bat.

For example, the height of the ball when the ball have travelled 1 m horizontally is:

1 + 1 - (1

The height of the ball when the ball have travelled 3 m horizontally is

1 + 3 - (3

If the ball hit the ground then h = 0m, right?aricho said:how far would the ball travel before it hits the ground?"

From there, for what x that makes h(x) = 0?

So when the ball hits the ground, how far has it travelled horizontally? Hint: you do know how to solve a

Can you go from here? :)

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