h ( t ) = − 16 t 2 + 72 t + 40 a) What is the maximum height of the ball? b) How many seconds does it take until the ball hits the ground?

Answer 1

The time when the ball hits the ground will be when h(t)=0; therefore, set h(t)=0 and solve the quadratic for t.

For a quadratic equation Ax² + Bx + C the maximum or minimum point will occur when

x=-B/2A. I cannot easily prove this to you but it is so; this fact will allow you to solve this problem.

Answer 2

h(t)=-16t^2+72t+40

a) this is a parabola that opens downward so it has a maximum point at (x,y) where x is the "maximizing" point and y is the maximum

x=-b/2a

x=-72/-32

x=2.25

h(2.25)=-16(2.25)^2+72*2.25+40

h(2.25)=-81+162+40

h(2.25)=121 feet is the maximum height of the ball

set h(t)=0

0=-16t^2+72t+40

divide all terms by 8

0=-2t^2+9t+5

multiply both sides by -1(all terms)

2t^2-9t-5=0

factor

(2t+1)(t-5)=0

disregard 2t+1=0 because it yields a negative number

t-5=0

t=5 seconds