To determine how long it takes for the object to hit the ground, we need to find the time [tex]\( t \)[/tex] when the object's height [tex]\( h \)[/tex] is zero. The equation for the height [tex]\( h \)[/tex] of the object is given by:
[tex]\[ h = -16t^2 + 96t + 256 \][/tex]
To find the time [tex]\( t \)[/tex] when the object hits the ground, we set [tex]\( h \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 96t + 256 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 96 \)[/tex]
- [tex]\( c = 256 \)[/tex]
We solve the quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 96^2 - 4(-16)(256) \][/tex]
[tex]\[ \Delta = 9216 + 16384 \][/tex]
[tex]\[ \Delta = 25600 \][/tex]
Now we apply the quadratic formula:
[tex]\[ t = \frac{-96 \pm \sqrt{25600}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-96 \pm 160}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-96 + 160}{-32} = \frac{64}{-32} = -2 \][/tex]
[tex]\[ t_2 = \frac{-96 - 160}{-32} = \frac{-256}{-32} = 8 \][/tex]
We need the physical solution for time, which cannot be negative. So, we discard [tex]\( t_1 = -2 \)[/tex] and take [tex]\( t_2 = 8 \)[/tex]:
Therefore, it takes [tex]\( 8 \)[/tex] seconds for the object to hit the ground.
