A ball is thrown vertically upward. After [tex]\( t \)[/tex] seconds, its height [tex]\( h \)[/tex] (in feet) is given by the function [tex]\( h(t) = 112t - 16t^2 \)[/tex]. After how long will it reach its maximum height?

Do not round your answer.

Time: [tex]\(\square\)[/tex] seconds



Answer :

To determine when the ball will reach its maximum height, we analyze the height function given:

[tex]\[ h(t) = 112t - 16t^2 \][/tex]

This is a quadratic function of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 112 \)[/tex]
- [tex]\( c = 0 \)[/tex] (though [tex]\( c \)[/tex] is not needed to find the maximum height).

For a quadratic equation [tex]\( at^2 + bt + c \)[/tex], the vertex (which gives the maximum or minimum value of the function) occurs at [tex]\( t = -\frac{b}{2a} \)[/tex]. Because the coefficient [tex]\( a \)[/tex] is negative in our case, the parabola opens downwards, and the vertex represents the maximum height.

Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

[tex]\[ t = -\frac{112}{2 \cdot -16} \][/tex]

First, calculate the denominator:

[tex]\[ 2 \cdot -16 = -32 \][/tex]

Then divide the numerator by the denominator:

[tex]\[ t = -\frac{112}{-32} = \frac{112}{32} \][/tex]

Simplifying this division:

[tex]\[ \frac{112}{32} = 3.5 \][/tex]

So, the ball reaches its maximum height at [tex]\( t = 3.5 \)[/tex] seconds.

Thus, the time at which the ball reaches its maximum height is:

[tex]\[ \boxed{3.5} \][/tex] seconds.