Quadratic equation question

Question

The height of an object t seconds after it is thrown from a height of h feet is modelled by the

equation h(t) = -16t^(2)+ vt + h. If the ball is thrown

from a point 6 feet above ground with an initial velocity, v, of 30 feet per second, how long will it take for the ball to hit the ground?

Answer 1

You are given the function h(t) = -16t² + vt + h for the height "h" as a function of time "t". If you look at the equation, you will see that when you plug in t = 0 that the height turns out to be "h". Since they tell you that the initial height is 6 feet, then the function must be h(t) = -16t² + vt + 6. You are also told that v = 30 therefore the function must be:

h(t) = -16t² + 30t + 6

The ball starts out (at t = 0) at 6 feet above the ground and will go up to a peak height then fall back to the ground. When it hits the ground, the height will be zero. So, the equation we must solve is:

0 = -16t² + 30t + 6

Start be dividing both sides by 2 to make the numbers a little easier to work with:

0 = -8t² + 15t + 3

Since this cannot be factored, use the quadratic formula:

So a = -8, b = 15, and c = 3:

t = [-15 ± √(15² - 4(-8)(3))]/[(2)(-8)

t = [-15 ± √321]/(-16)

t = (-15 + 17.916)/(-16) and t = (-15 - 17.916)/(-16) or t = -0.18 and t = 2.06

We can throw out the negative answer since we will not be considering answers that are a negative time. So the object takes approx 2.06 seconds to hit the ground.

Answer 2

0 = -16t² + 30t + 6

Can you factor (like in Algebra 1), use the Zero Product Rule, and answer?