Quadratic Equation/Algebra

Question

If football is punted into the air. It’s height h, in meters, after t seconds is given by the equation h=-4.9+24.5t+1. How high is the ball after 1 second? Find the maximum height of the ball to one decimal place?

Answer 1

h(t) = (a/2)t^2 +vot + ho

where a = acceleration=-4.9 meters per second per second = -4.9 m/sec^2 due to gravity, vo= initial velocity=24.5 m/sec, ho= initial height = 1 meter

h(t) = -4.9t^2 +24.5t + 1

h(1) = -4.9 +24.5 +1 = 25.5-4.9 = 20.6 meters high after 1 second

max height is when derivative = 0

h'(t) = -9.8t +24.5 = 0

t = 24.5/9.8

max height = h(24.5/9.8) = -4.9(24.5/9)^2 +24.5(24.5/9) +1

= -4.9(600.25/96.04) + 600.25/9 +1

= -600.25/19.6 +600.25/9 + 1

= -30.625 +67.69444...

= 37.0694444...

= about 37.1 meters = max height