Answer:
It takes 4 seconds for the projectile to hit the ground
Step-by-step explanation:
The height of the projectile after t seconds is given by the following equation:
[tex]h(t) = -16t^{2} + 32t + 128[/tex]
How long will it take the projectile to hit the ground?
It happens when [tex]h(t) = 0[/tex]
So
[tex]h(t) = -16t^{2} + 32t + 128[/tex]
[tex]-16t^{2} + 32t + 128 = 0[/tex]
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this question:
[tex]-16t^{2} + 32t + 128 = 0[/tex]
So [tex]a = -16, b = 32, c = 128[/tex]
[tex]\bigtriangleup = 32^{2} - 4*(-16)*(128) = 9216[/tex]
[tex]t_{1} = \frac{-32 + \sqrt{9216}}{2*(-16)} = -2[/tex]
[tex]t_{2} = \frac{-32 - \sqrt{9216}}{2*(-16)} = 4[/tex]
Time is a positive measure, so:
It takes 4 seconds for the projectile to hit the ground
