We will determine the maximum height of the baseball as follows:
We will need the following formulas:
[tex]v=u+at[/tex][tex]s=ut+\frac{1}{2}at^2[/tex]Here "u" represents the original speed, "t" represents the time, "a" the acceleration of the body and "s" is the total discance moved. [We will find s to solve the problem].
So:
First we have that the acceleation that the body will experience is -9.8m/s^2 [Since the object is going upwards and gravity is pulling on it towards the ground]. [acceleration of gravity using feet over second squared is 32.17 ft/s^2].
[tex]v=(12.4968)+(-9.8)t[/tex]But the maximum height will be reached when the velocity after certain time has passed is 0 ft/s, so:
[tex]0=(12.4968)+(-9.8)t\Rightarrow9.8t=12.4968[/tex][tex]\Rightarrow t=1.275183673\ldots\Rightarrow t\approx1.3[/tex]So, at approximately 1.3 seconds the maximum heigth willl be reached.
Now, we solve for s:
[tex]s=(12.4968)(1.275183673)+\frac{1}{2}(-9.8)(1.275183673)^2\Rightarrow s=7.967857665[/tex][tex]\Rightarrow s\approx8[/tex]So, the maxumum altitude for the baseball will be 8 meters, but we have to add the initial 5 feet at which it was launched:
[tex]h\approx8+1.524\Rightarrow h\approx9.491857665[/tex]And taking that to feet we will have:
[tex]h\approx31.141265305118\ldots[/tex]So, the solution must be the last option. [The discrepanc