A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in meters t seconds after it has been thrown. Find the maximum height reached by the ball and the time at which that happens.

Now we can set up the derivate equal to 0 and we have:

And solving for t we got:

For the second derivate respect the time we got:

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

In order to maximize this function we need to take the first derivate respect the time and we have:

Now we can set up the derivate equal to 0 and we have:

And solving for t we got:

For the second derivate respect the time we got:

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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