MHB Application of quadratic function in kinematics

AI Thread Summary
The discussion focuses on evaluating the quadratic function h(t) = -16t^2 + 80t, which models the height of a golf ball over time. The participant's answers for evaluating h(4), explaining its meaning, determining maximum height, and the time of maximum height were confirmed as correct. The maximum height of the golf ball is 100 feet, occurring at 2.5 seconds after being struck. Several methods to find the maximum height and axis of symmetry were discussed, including factoring, using the vertex formula, and completing the square. Overall, the application of quadratic functions in kinematics is effectively illustrated through this golf ball example.
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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
 
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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:

$$h(t)=80t-16t^2=16t(5-t)$$

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

$$t=\frac{0+5}{2}=\frac{5}{2}$$

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

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

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

$$x=-\frac{b}{2a}$$

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

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

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:

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

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