To determine when the object will hit the ground, we need to find when the height [tex]\( h \)[/tex] is equal to 0. The height [tex]\( h \)[/tex] of the object at any time [tex]\( t \)[/tex] is given by the quadratic equation:
[tex]\[ h = -16t^2 + 48t + 160 \][/tex]
We set [tex]\( h \)[/tex] to 0 to find the time [tex]\( t \)[/tex] when the object hits the ground:
[tex]\[ 0 = -16t^2 + 48t + 160 \][/tex]
This is a quadratic equation in the standard form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 160 \)[/tex]. We need to solve this quadratic equation for [tex]\( t \)[/tex].
The solutions to the quadratic equation [tex]\( at^2 + bt + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our case:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 160 \)[/tex]
Plugging these values into the quadratic formula, we get:
[tex]\[ t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(160)}}{2(-16)} \][/tex]
Now, we calculate the discriminant [tex]\( \Delta \)[/tex] first:
[tex]\[ \Delta = 48^2 - 4(-16)(160) = 2304 + 10240 = 12544 \][/tex]
Then we calculate the square root of the discriminant:
[tex]\[ \sqrt{12544} = 112 \][/tex]
Now we can find the values of [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-48 \pm 112}{-32} \][/tex]
This gives us two potential solutions:
1. [tex]\( t_1 = \frac{-48 + 112}{-32} = \frac{64}{-32} = -2 \)[/tex]
2. [tex]\( t_2 = \frac{-48 - 112}{-32} = \frac{-160}{-32} = 5 \)[/tex]
Hence, the solutions to the quadratic equation are [tex]\( t = -2 \)[/tex] and [tex]\( t = 5 \)[/tex].
The negative value [tex]\( t = -2 \)[/tex] does not make sense in the context of time since time cannot be negative. Therefore, we discard it.
The object will hit the ground after [tex]\( t = 5 \)[/tex] seconds.
