Quadratic problem

  • Thread starter aricho
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  • #1
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" 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. how far would the ball travel before it hits the ground?"

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
 

Answers and Replies

  • #2
Integral
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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
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Why Pythagoras? Do you have a right triangle here? Do you even have straight lines here?
 
  • #4
BobG
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I have to admit, considering you titled the problem as a quadratic problem, the choice of the Pythagorean Theorem was kind of interesting. The quadratic equation would work better. :biggrin:
 
  • #5
VietDao29
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As others have pointed out that you should solve h = 0 for x.
I just want to make it a little bit clearer...
aricho 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.
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 - (x2 / 40).
For example, the height of the ball when the ball have travelled 1 m horizontally is:
1 + 1 - (12 / 40) = 2 - 1 / 40 = 79 / 40 (m).
The height of the ball when the ball have travelled 3 m horizontally is
1 + 3 - (32 / 40) = 4 - 9 / 40 = 151 / 40 (m), do you get it?
aricho said:
how far would the ball travel before it hits the ground?"
If the ball hit the ground then h = 0m, right?
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 Quadratic equation, right?
Can you go from here? :)
 

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