Question

5. A ball is kicked into the air and follows the path described by h(t)= -4.9t^2+6t+0.6. where t is the time in seconds and h is the height in meters above the ground. Find the maximum height of the ball. What value would you have to change in the equation if the maximum height of the ball is more than 2.4 meters?

Answer 1

Find how long the ball will be in the air by solving for h = 0
-16t^2 + 80t = 0
Factor out -t:
-t(16t - 80) = 0
Two solutions:
-t = 0
and
16t = +80
t = 80/16
t = 5 secs for the ball to hit the ground
:
You can find the time of max height by finding the axis of symmetry:
Using the form: y = ax^2 + bx + c; the axis of symmetry: x = -b/(2a)
:
Putting the equation in this form, we have: h = -16t^2 + 80t: a=-16, b = 80
t = -80/(2*-16)
t = -80/-32
t = +2.5 seconds to reach max height
:
We find the max height by substituting 2.5 for t in the original equation
h = 80(2.5) - 16(2.5^2)
h = 200 - 16(6.25)
h = 200 - 100
h = 100 ft is the max height
:
We find the time for 60ft of height by substituting 60 for h in the original equation
60 = 80t - 16t^2
arrange a quadratic equation on the left
16t^2 - 80t + 60 = 0
Use the quadratic formula to solve for t:

:

:

:
Solve this and you will have two solutions of:
t = .92 sec, 60 ft on the way up
t = 4.08 sec, 60 ft on the way down