<PRE><B>Question 1107698</B>
The equation of the path of a cricket ball thrown at an angle of 45 degrees, with the horizontal is y = x - (x^2)/50 where x metres and y metres are the horizontal distance traveled and vertical height respectively. Calculate the greatest vertical height reached and the total horizontal distance traveled. 
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Max occurs when 1-(2/50)x = 0
(1/25)x = 1
x = 25 metres
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max height occurs at f(25)= [25 - (25^2)/50]= 25 - (625/50) = 25-12.5 = 12.5 metres
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Cheers,
Stan H.
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