SOLUTION: A stone is thrown directly upward from a height of 50 feet with an initial velocity of 56 ft/sec. The height of the stone "t" seconds after it has been thrown is given by the formu

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A stone is thrown directly upward from a height of 50 feet with an initial velocity of 56 ft/sec. The height of the stone "t" seconds after it has been thrown is given by the formu      Log On


   



Question 806717: A stone is thrown directly upward from a height of 50 feet with an initial velocity of 56 ft/sec. The height of the stone "t" seconds after it has been thrown is given by the formula s(t)=-16t^2+56t+50. Use the appropriate algebraic formula to determine the time at which the ball reaches its maximum height and find what the maximum height is.
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
s(t)=-16t^2+56t+50
parabola
focus | (7/4, 6335/64)=(1.75, 98.9844)
vertex | (7/4, 99)~~(1.75, 99)
semi-axis length | 1/64 = 0.015625
focal parameter | 1/32 = 0.03125
eccentricity | 1
directrix | s = 6337/64
vertex is the high point
98.9844 ft at 1.75 seconds