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.
[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.