SOLUTION: The equation of the path of a cricket ball thrown at an angle of 45 degrees to the horizontal is y = x - x^2/50, where x and y are the horizontal distance travelled and the vertica

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The equation of the path of a cricket ball thrown at an angle of 45 degrees to the horizontal is y = x - x^2/50, where x and y are the horizontal distance travelled and the vertica      Log On


   

Question 1011969: The equation of the path of a cricket ball thrown at an angle of 45 degrees to the horizontal is y = x - x^2/50, where x and y are the horizontal distance travelled and the vertical height respectively. Calculate the greatest vertical height reached and the horizontal distance travelled.
What i have done so far:
y = x - x^2/50
50y = 50x - x^2
= -(x^2 - 50x) + 0
= - (x-25)^2 + 625
Therefore greatest height would be 625 divided by 50 which is 12.5, and that answer is correct.
But I don't know how to get horizontal distance???
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
When the height (y)=0, the ball is at minimum or maximum distance
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y=x-x%5E2%2F50
Let y=0:
0=x-x%5E2%2F50
0=50x-x%5E2
0=%28x%29%2850-x%29
x=0 OR 50-x=0
x=0 OR 50=x
The minimum horizontal distance (start) is 0,
the maximum horizontal distance (ball hits ground) is 50.