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 since the ball was thrown upward from a bridge that is same distance above the ground, Let f(t) = -16t^2 + 46t +259. Approximately how high is the bridge above the ground? After how many seconds does the ball list the ground? After how many seconds does the ball reach its maximum height above the ground? What is the maximum height above the ground reached by the ball?

Solution

f(t)=-16t²+46t+259

a)before the ball is thrown , it was on bridge

=>at t =0 ball is on bridge

height of bridge above the ground =f(0)

height of bridge above the ground =259 ft

b) when ball hits the ground ,height is 0

=> f(t)=0

=>-16t²+46t+259 =0

by quadratic formula

=>t =5.71

after 5.71 seconds ball hits the ground

c)f(t)=-16t²+46t+259

f(t)=-16(t²-(46/16)t) +259

f(t)=-16(t²-(46/16)t+(23/16)²-(23/16)²) +259

f(t)=-16(t²-(46/16)t+(23/16)²)+(23²/16) +259

f(t)=-16(t-(23/16))²+(4673/16)

vertex of parabola is (23/16,4673/16)

ball reach maximum height when t =23/16 sec

ball reach maximum height when t =1.4375 sec

d) maximum height =4673/16 feet

maximum height =292.0625 feet