Soccer ball Height

Question

1. The path of a soccer ball can be given by the quadratic function: h=29.4t-4.9t^2 where t s the time in seconds and h is the height in meters. (5 marks)
2. Put the function in vertex form, and state the maximum height reached by the ball.
3. How many seconds did it take to reach this height?

Please answer in full detail

Answer 1

Since it appears this question may be a part of an exam, I'll show you how the problem is done using a similar problem.

Let the path of a soccer ball can be given by the quadratic function: h = 96t - 16t² where t is the time in seconds and h is the height in feet.

Vertex form is: h(t) = a(t - h)² + k where the vertex is the point (h, k). To transform the equation h = 96t - 16t² into that form you can either "complete the square" or you can find the vertex and then plug it into the format. I'll do it both ways:

1) Complete the square:

a) Write the equation in standard form: h(t) = -16t² + 96t

b) Factor out the coefficient in front of the t², which is -16: h(t) = -16(t² - 6t)

c) Using the coefficient in front of the "t" (which is -6), take half of it (making it -3) and then square it (making it 9). Add in the "9" but subtract what you have added to make the change "net zero". In this case, there is a factor in front of the quadratic of "-16" so when adding in "9" inside the parenthesis, you have actually added (-16)•9 or -144. So to make a net zero, you must add 144 to the entire equation making:

h(t) = -16(t² - 6t + 9) + 144

d) Write the quadratic as a square: h(t) = -16(t - 3)² + 144

The vertex then is at (3, 144) so the maximum height is 144 feet at that occurs after 3 seconds.

2) Finding the Vertex and then Creating the Vertex Form:

The "t-coordinate" of the vertex is found at t = -b/(2a) for h(t) = at² + bt + c. In the case of h(t) = 96t - 16t², "a = -16" and "b = 96" so t = (-96)/(2•(-16)) = -96/-32 = 3 seconds. The value of the function at t = 3 is found by plugging in t = 3: So h(3) = 96(3) - 16(3)² = 288 - 144 = 144 feet. So the vertex is (3, 144). Working backwards, by plugging these into the vertex form h(t) = a(t - h)² + k we get h(t) = a(t - 3)² + 144 and comparing this to the original h(t) = 96t - 16t² it appears a = -16. Plugging that in we get h(t) = -16(t - 3)² + 144 (which matches what we got when we completed the square). To make sure this is correct, multiply it out and compare to the original. It does turn out to be the h(t) = 96t - 16t² when multiplied out.

Answer 2

Answer 3

h(t) = -4.9t^2 + 29.4t

take the derivative, and set=0

h'(t) =-9.8t +29.4 = 0

t = 29.4/9.8

= 3 seconds to reach max height

or find -b/2a where h(t) = ax^2 +bx +c, with a=-4.9, b=29.4, c=0

-29/2(-4.9) = 3

to find maximum height plug t=3 into the original equation:

max height = h(3) =-4.9(3)^2+29.4(3)

= 88.2 - 44.1

= 44.1 meters

the height equation graphically describes a downward opening parabola

with vertex = (t,h) = (3, 44.1)=maximum point

graphically, quadratic functions are parabolas.

if the squared term has a negative coefficient, it's a downward opening parabola

h(t) = h = y = -4.9t^2 +29.4t = -4.9(t^2 - 3t) = -4.9t(t-6)

t or x intercepts or zeros are 0 and 6.

vertex has an x or t coordinate midway between 0 and 6,

(0+6)/2 = 3

h(t)= -4.9t^2 +29.4t

= -4.9(t^2 -6t)

complete the square, add and subtract 9(4.9)

=-4.9(t^2 -6t + 3^2) + (3^2)(4.9)

= -4.9(t-3)^2 +4.9(9)

=-4.9(t-3)^2 + 44.1 in vertex form

vertex form is

y(t)= -(a/2)(t-h)^2 + k where (h,k)= vertex

in this specific example h=3 and k=44.1

vertex = (h,k) = (3, 44.1)

OR

you can just go to an online graphing calculator

or a handheld graphing calculator

It will graph the function for you

and you can read the vertex off the graph

just

plug in

-4.9x^2 + 29.4x

it will show you a graph with max point = (3,44.1)

that's the vertex with x or t =3 seconds

y or h(t) or f(x) = 44.1 meters

the given equation h(t) = -4.9t^2 + 29.4t is actually a close approximation to the effect of gravity at sea level

h(t) = -(9.80655/2)t^2 +29.4t is a closer approximation

using ht(t) =-4.903275t^2 + 29.4t will give a better more accurte solution,although they're very similar.

work out the arithmetic

or plug that quadratic into an online graphing calculator such as Desmos

knowing altitude and air resistance also alter the solution slightly, but not much

then there's the force of boyancy, the relative density of the air compared to the density of the air filled soccer ball. If this ball were thrown up in a super smogged filled area, such as Mexico City or a city in China on a particularly pollution smoke filled day, it won't go nearly as high, and it's travel time will be much longer. Ask this question in an environmentally friendly academic course, and the answer will be substantially different. if the smog is dense enough the soccer ball will float away like a helium filled balloon and never come down.