Question

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.

Answer 1

Answer:

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

Step-by-step explanation:

The height of the ball after t seconds can be modeled with equation

$h(t)=104t-16t^2$

The graph of this quadratic function is parabola which opens downwards.

Find the coordinates of the parabola:

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

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