To determine how many seconds it will take for the frog to land on the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is 0. The height function given is:
[tex]\[ h(t) = -16t^2 + 64t + 3 \][/tex]
We set [tex]\( h(t) \)[/tex] to 0 to find the times when the frog lands on the ground:
[tex]\[ 0 = -16t^2 + 64t + 3 \][/tex]
This is a quadratic equation in the form of:
[tex]\[ 0 = at^2 + bt + c \][/tex]
Where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 64 \)[/tex]
- [tex]\( c = 3 \)[/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-64 \pm \sqrt{64^2 - 4(-16)(3)}}{2(-16)} \][/tex]
First, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 64^2 - 4 \cdot (-16) \cdot 3 \][/tex]
[tex]\[ \Delta = 4096 + 192 \][/tex]
[tex]\[ \Delta = 4288 \][/tex]
Next, we substitute the discriminant back into the quadratic formula:
[tex]\[ t = \frac{-64 \pm \sqrt{4288}}{-32} \][/tex]
Simplifying the square root of the discriminant:
[tex]\[ \sqrt{4288} = \sqrt{64 \cdot 67} = 8\sqrt{67} \][/tex]
Substituting this back into the equation:
[tex]\[ t = \frac{-64 \pm 8\sqrt{67}}{-32} \][/tex]
We simplify by dividing the entire numerator and the denominator by -8:
[tex]\[ t = \frac{64 \mp 8\sqrt{67}}{32} \][/tex]
[tex]\[ t = 2 \mp \frac{\sqrt{67}}{4} \][/tex]
Therefore, the two solutions for [tex]\( t \)[/tex] are:
[tex]\[ t_1 = 2 - \frac{\sqrt{67}}{4} \][/tex]
[tex]\[ t_2 = 2 + \frac{\sqrt{67}}{4} \][/tex]
These solutions represent the two times when the frog is at ground level. The negative time is not physically meaningful in this context, so we discard it.
Hence, the correct time for the frog to land on the ground is:
[tex]\[ t = 2 + \frac{\sqrt{67}}{4} \][/tex]
So, it will take approximately [tex]\( 2 + \frac{\sqrt{67}}{4} \)[/tex] seconds for the frog to land on the ground.
