A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=108t-16t^2 . What is the maximum height that the ball will reach?
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Respuesta :

h(t)= 108t-16t^2
y=mx+c
y= 182.25 ft
h(t)= 182.25ft

The maximum height that the ball will reach should be 18.25 feet.

Given information:

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function [tex]h(t) = 108t - 16t^2[/tex]

Calculation of maximum height:

Since [tex]h(t) = 108t - 16t^2[/tex]

So here we applied y=mx+c

So,

y= 182.25 ft

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