A ball is thrown vertically upward. After seconds, its height (in feet) is given by the function h(t)=104t-16t^2 . After how long will it reach its maximum height?
Do not round your answer.

[tex]t_v=\dfrac{-b}{2a}=\dfrac{-104}{2\cdot (-16)}=\dfrac{13}{4}=3\dfrac{1}{4}\ seconds\\ \\h(t_v)=104\cdot \dfrac{13}{4}-16\cdot \left(\dfrac{13}{4}\right)^2=26\cdot 13-13^2=169\ feet[/tex]

The maximum height is 169 feet and the ball will reach this maximum height after [tex]3\frac{1}{4}[/tex] seconds.

abidemiokin abidemiokin · Answer 2 · 2022-04-01T13:06:59+02:00

It will take the ball 3.25 seconds for it to reach its maximum height.

Functions and derivative

Given the height of the ball launched expressed as h(t)=104t-16t²

The velocity of the ball at the maximum height is zero, hence;

[tex]\frac{dh}{dt}=104-32t[/tex]

Since dh/dt = 0, hence;

[tex]0=104-32t\\32t=104\\t=3.25secs[/tex]

Therefore it will take the ball 3.25 seconds for it to reach its maximum height.

Learn more functions and derivative here; https://brainly.com/question/12047216

A ball is thrown vertically upward. After seconds, its height (in feet) is given by the function h(t)=104t-16t^2 . After how long will it reach its maximum height? Do not round your answer.

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Functions and derivative

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A ball is thrown vertically upward. After seconds, its height (in feet) is given by the function h(t)=104t-16t^2 . After how long will it reach its maximum height?
Do not round your answer.