The path of a basketball thrown at an angle of 45 degreess can be measured by the equation :
y=-0.02x^2 + x + 6
where y is the vertical distance and x is the horizontal distance
a. Find the maximum height of the basketball
b. How far does the basketball travel when it reaches its maximum height?
+130536 y= -0.02x^2 + x + 6
Answering rhe second question, first.....the horizontal distance traveled =
-1/[2(-.02)] = 25 ft (???)
And the max ht. is given by
-.02(25)^2 + 25 + 6 = 18.5 (ft ??)
Here's a graph of the function........https://www.desmos.com/calculator/mz7afyqedt
