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Application of quadratic function in kinematics


New member
Aug 7, 2013
Basically I don't know anyone in real life that can help me with this, so I need help checking to see if my answers are correct :)

Part B

1) A golfer hits a nice iron shot and the ball's height above the ground is given by h(t) = -16t^2 + 80t, where t is the time in seconds since the ball was hit.

a) Evaluate h(4)

My Answer: 64

b) Explain the meaning of evaluating h(4) in the context of the problem.

My Answer: Evaluating h(4) will find the height in feet of the golf ball when it is 4 seconds after being struck by the golfer.

c) Determine the max height the golf ball reaches during the shot

My Answer: 100 feet

d) Determine the time, t, when the maximum occurs.

My Answer: 2.5 seconds


Staff member
Feb 24, 2012
Re: Please check my answers - 7

All correct.

As far as the maximum is concerned, there are several ways we can do this, and this doesn't even include using the calculus! :D

We know this parabola opens down, as the leading coefficient is negative. Thus, its global maximum will be at the vertex.

i) Factor:

\(\displaystyle h(t)=80t-16t^2=16t(5-t)\)

We see the roots are at \(\displaystyle t=0,\,5\) and so the axis of symmetry, that value of $t$ on which the vertex lies, must be midway between the roots:

\(\displaystyle t=\frac{0+5}{2}=\frac{5}{2}\)

\(\displaystyle h\left(\frac{5}{2} \right)=16\cdot\frac{5}{2}\left(5-\frac{5}{2} \right)=(2\cdot5)^2=100\)

ii) Find the axis of symmetry without using the roots:

A quadratic of the form \(\displaystyle y=ax^2+bx+c\) will have an axis of symmetry given by:

\(\displaystyle x=-\frac{b}{2a}\)

and so for the given function, we find the axis of symmetry is:

\(\displaystyle t=-\frac{80}{2(-16)}=\frac{5}{2}\)

Finding the value of the function at this value of $t$ is the same as above.

iii) Write the function in vertex form:

Completing the square, we find:

\(\displaystyle h(t)=-16t^2+80t=-16\left(t^2-5t+\left(\frac{5}{2} \right)^2 \right)+16\left(\frac{5}{2} \right)^2=-16\left(t-\frac{5}{2} \right)^2+100\)

And so we find the vertex is at \(\displaystyle \left(\frac{5}{2},100 \right)\)