Question e8c1c

Jun 13, 2017

1) $t \approx 3.87$ seconds
2) $\text{H"=42"ft}$

Explanation:

1) If I'm correct, to solve the first question, we set $\text{H}$ equal to $0$ because $\text{H}$ is the height of the ball and we are looking for after how many seconds $t$ will the ball hit the ground. Well, we know that the ball will be on the ground when the ball is at a height of $0$.

We set up an equation like this and solve for $t$:

$- 16 {t}^{2} + 60 t + 6 = 0$

You may have noticed that this isn't easy to solve by factoring but we can find the answer by using the quadratic formula:

$t = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We let $a = - 16 , b = 60 , c = 6$ and substitute those values into the quadratic formula and solve:

 t = (-(60) \pm sqrt((60)^2-4(-16)(6))) / (2(-16) #

$t = \frac{- 60 \setminus \pm \sqrt{249}}{- 32}$

$t = \frac{15 \setminus \pm \sqrt{249}}{8}$

If we evaluate this into a calculator we will get two answers:

$t \approx 3.87 , t \approx - 0.09$

We can ignore the second answer because since we are talking about time, time can't be negative if we are initially starting when $t = 0$ seconds. Therefore, the ball will hit the ground at approximately $3.87$ seconds ($4$ seconds if you want to be loose about it)

2) For this question we simply substitute $3$ for $t$ to find the height ($\text{H}$). Thus,

$\text{H} = - 16 {\left(3\right)}^{2} + 60 \left(3\right) + 6$

After some algebra...

$\text{H"=42 "ft}$ (after $3$ seconds)

P.S. It's good to always visualize the problem. Below is what the path of the ball looked like given by the function in the problem:

graph{-16x^2+60x+6 [-6.7, 153.3, -16.6, 63.4]}