Answer:
t = 2.11 seconds
Step-by-step explanation:
A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:
[tex]h=-16t^2+32t+48[/tex]
Let us assume that we need to find the time by it to reach the ground. It means h = 0
[tex]-16t^2+32t+48=0[/tex]
The above is a quadratic equation. The value of t is given by :
[tex]t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s[/tex]
So, it will take 2.11 seconds to reach the ground.
