
Jordan K. answered 10/26/15
Tutor
4.9
(79)
Nationally Certified Math Teacher (grades 6 through 12)
Hi Haley,
Let's begin by writing the quadratic function in standard form
(ax2 + bx + c):
h(t) = -16t2 + 40t + 5 (a = -16; b = 40; c =5)
Next, let's find the time (t) needed for the ball to reach its maximum height using the axis of symmetry formula (x = -b/2a) for a parabola:
t = -40/[2(-16)]
t = 40/32
t = 5/4
t = 1.25 seconds to reach maximum height
Next, let's plug in the value of 1.25 for t into the height function, h(t), to calculate the maximum height of the ball:
h(t) = -16(1.25)2 + 40(1.25) + 5
h(t) = -16(1.5625) + 50 + 5
h(t) = -25 + 55
h(t) = 30 feet (maximum height)
Finally, let's set the height function, h(t), to zero and determine the time (t) when the ball hits the ground:
-16t2 + 40t + 5 = 0
-16t2 + 40t = -5
(-16t2)/(-16) + 40t/(-16) = -5/(-16)
t2 - 2.5t = 0.3125
t2 - 2.5t + [(0.5)(2.5)]2 = 0.3125 + [(0.5)(2.5)]2
t2 - 2.5t + (1.25)2 = 0.3125 + (1.25)2
t2 - 2.5t + 1.5625 = 0.3125 + 1.5625
(t - 1.25)2 = 1.875
t - 1.25 = +/- √1.875
t = 1.25 +/- 1.3693
t = 1.25 + 1.3693 (only accept positive root)
t = 2.6193 seconds to hit the ground
Below is the link to our graph of the height function, h(t):
https://dl.dropbox.com/s/5riq3ijc7jdu2j3/Graph_of_Height_Function_for_Thrown_Ball.png?raw=1
The graph shows three points for the height function, h(t):
1. y-intercept (initial height of the ball, i.e. the height from which the ball is thrown) - when t = 0, the value of h(t) is 5 (the constant term of the function).
2. vertex (maximum height of the ball) - also determined algebraically above.
3. positive root or zero (time at which the ball hits the ground) - also determined algebraically above.
Thanks for submitting this problem and glad to help.
God bless, Jordan (Romans 5:8)