Question

Kyle throws a baseball straight up from a height of 5 feet with an initial speed of 41feet per second.What is the maximum height of the baseball?

Answer 1

We will determine the maximum height of the baseball as follows:

We will need the following formulas:

`v=u+at`
`s=ut+(1)/(2)at^2`

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].

`v=(12.4968)+(-9.8)t`

But the maximum height will be reached when the velocity after certain time has passed is 0 ft/s, so:

`0=(12.4968)+(-9.8)t\Rightarrow9.8t=12.4968`
`\Rightarrow t=1.275183673\ldots\Rightarrow t\approx1.3`

So, at approximately 1.3 seconds the maximum heigth willl be reached.

Now, we solve for s:

`s=(12.4968)(1.275183673)+(1)/(2)(-9.8)(1.275183673)^2\Rightarrow s=7.967857665`
`\Rightarrow s\approx8`

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:

`h\approx8+1.524\Rightarrow h\approx9.491857665`

And taking that to feet we will have:

`h\approx31.141265305118\ldots`

So, the solution must be the last option. [The discrepanc