Question

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 h=-16t^2+32t+48

Answer 1

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:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$

So, it will take 2.11 seconds to reach the ground.