SOLUTION: During a halftime show, a baton twirler releases her baton from a point 4 feet above the ground with an initial vertical velocity of 25 feet per second. Use a vertical motion model

Question 174670This question is from textbook Algebra 2, Texas Edition
: During a halftime show, a baton twirler releases her baton from a point 4 feet above the ground with an initial vertical velocity of 25 feet per second. Use a vertical motion model to write a function for the height h (in feet) of the baton after t seconds then graph the function and label the vertex.
Please show all work possible and any strategies used in depth. Thank You for your time and help! =) Hope to hear from you soon. This question is from textbook Algebra 2, Texas Edition

Found 2 solutions by Earlsdon, josmiceli:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
You can use the function for the height of an object propelled upwards with an initial velocity of from an initial height of :

In this problem, and , so, making the substitutions, we get:

You can find the t-value of the vertex by:
where: b = 25 and a = -16, so:

Let's look at the graph of this function:

The vertex is the point at the top of the curve.
Since we already have the t-value of the vertex, we can find the h-value by substituting into the above equation and we get
So the vertex is located at (0.78, 13.77)
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
For motion in the vertical direction, the acceleration due
to gravity has to be included in the equation of motion
Motion in the horizontal direction is not included in this
problem.
The equation for motion in the vertical direction (on planet earth)is

Where is the initial velocity of the baton, so

There is an initial height also, of 4 ft, so the equation becomes

Here's a plot. is the vertical axis and is the horizontal axis

The graph shows the situation correctly
At , as it should
The baton takes the same amount of time to go from to it's
peak as it does to go from the peak back to . Then it continues
on to the ground.
-----------------------
The vertex is midway between the roots of the equation. You can
see by the graph that one root is negative and one is positive. the
vertex ends up being at when the equation is in the
form . In our equation

This is the time in seconds that it takes to reach the vertex
Now plug this back into the equation

ft
This is verified by the graph, but you didn't really
need this- it's extra info.

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