The function f gives the height of a ball h above the ground

The function f gives the height of a ball, h, above the ground (measured in feet) in terms of the number of seconds elapsed since the ball was thrown upward from a bridge that is some distance above the ground, t. Let h = f(t) and f(t) = - 16t^2 + 20t + 206. a. Approximately how high is the bridge above the ground? b. After how many seconds does the ball hit the ground? seconds after the ball was thrown c. After how many seconds does the ball reach its maximum height above the ground? d. What is the maximum height above the ground reached by the ball?

Solution

f(t) = -16t^2 + 20t + 206

a) Bridge is 206 ft above ground

b) Ball hits the ground:

f(t) = 0 ; -16t^2 + 20t + 206 =0 ; solve for t

t = -3.017 , 4.267 .Neglect -ve t

So, solution t = 4.267sec

c)Ball reaches maximum at vertex t = -b/2a = -(20/2*-16) = 0.625

d) Maximum height plug t = 0.625 in f(t) :

f(0.625) = -16(0.625)^2 + 20(0.625) + 206 = 212.25 ft