F a baseball is projected upward from ground level with an initial velocity of 96 feet per second, then its height is a function of time, given by sequalsminus16tsquaredplus96t. what is the maximum height reached by the ball?

here in order to find the maximum height we will use the concept of calculus where the maximum value of function will exist when its derivative becomes zero.

so here we can say

[tex]\frac{dh}{dt} = 0[/tex]

[tex]-32 t + 96 = 0[/tex]

[tex]t = 3 s[/tex]

now we will put t = 3 in the given equation

[tex]h = -16*(3)^2 + 96*(3)[/tex]

[tex]h = 144 ft[/tex]

So the maximum height will be h = 144 ft.

F a baseball is projected upward from ground level with an initial velocity of 96 feet per​ second, then its height is a function of ​time, given by sequalsminus16tsquaredplus96t. what is the maximum height reached by the​ ball?

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F a baseball is projected upward from ground level with an initial velocity of 96 feet per second, then its height is a function of time, given by sequalsminus16tsquaredplus96t. what is the maximum height reached by the ball?