Part 1:
Replace "v0" in the given equation with the given velocity of 15 ft/sec:
y = -16t^2 + 15t +6.5
Part 2:
Now set the equation to 0 which would be when the basketball hits the ground:
-16t^2 +15t +6.5 = 0
A quadratic equation is solved using the formula:
t = -b +/- sqrt(b^2-4ac)/2a
Using the given equation: a = -16, b = 15 and c = 6.5
Replace the values and solve:
t = -(15)+/- sqrt(15^2 -4(-16)(6.5))/2(-16)
This solves to get both -0.32 and 1.26 seconds.
The time has to be a positive value so t = 1.26 seconds.
Part3:
Using the quadratic form at^2 + bt + c
The maximum is found using t = -b/2a = -15/2(-16) = = 0.47 seconds
The maximum height would be at 0.47 seconds
Part 4:
Replace t with 0.47 and solve for maximum height:
y = -16(0.47)^2 + 15(0.47) +6.5
Maximum height would be 3.52 feet
