William W. answered • 02/02/20
Math and science made easy - learn from a retired engineer
You can use desmos dot com to graph h = −16t2 + 25t + 5 and you will get this graph:
It shows that the ball goes up to a maximum height of just a little higher than 14 feet and then beck down to a height of zero at a time just greater than 1.7 seconds. If you wanted to graph it manually (paper/pencil), then you would build a table. You would pick values of "t" and then plug those values into the equation to get a corresponding value of "h" like this:
t | h
-------|-----
0 | 5
0.2 | 9.36
0.4 | 12.44
0.6 | 14.24
0.8 | 14.76
1.0 | 14
1.2 | 11.96
1.4 | 8.64
1.6 | 4.04
1.7 | 1.26
You can now graph these ordered pairs and connect up the points with a curved line.
Now, since you are interested in what happens at the maximum and minimum heights, you can focus on times that appear to be close to those and pick additional numbers around those values.
From the tale above, it appears the maximum occurs near t = 0.8 s so let's pick numbers on both sides of that value to see what is going on:
t | h
-------|-----
0.79 | 14.7644
0.81 | 14.7524
The numbers show the ball is lower at 0.81 s than it is at 0.79 s so let's look for the maximum might before 0.79 s:
t | h
-------|-----
0.77 | 14.7636
0.78 | 14.7656
It appears that at 0.78 s the ball reaches a maximum height of about 14.77 feet
Now, lets look at the minimum height. Obviously that must be at height 0 (ground level). The first table shows this occurs just a little after 1.7 seconds. Lets focus in on that time:
t | h
-------|-----
1.71 | 0.9644
1.72 | 0.6656
1.73 | 0.3636
1.74 | 0.0584
1.75 | -0.25
Obviously the last entry that results in a negative height can't happen. So it appears that the minimum occurs at about 1.74 s and is 0 feet.
To do all this algebraically, requires skills typically taught in Algebra 2 so I'll assume the technique described above is what you are supposed to do. If you need to solve it algebraically, please comment that to me and I'll help you with that.
