Question

Question #e8c1c

Answer 1

1) `t~~3.87` seconds
2) `"H"=42"ft"`

Explanation:

1) If I'm correct, to solve the first question, we set `"H"` equal to `0` because `"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`:

`-16t^2+60t+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 = (-b \pm sqrt(b^2-4ac)) / (2a)`

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 = (-60 \pm sqrt(249)) / (-32)`

`t = (15 \pm sqrt(249)) / (8)`

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

`t~~3.87, t~~-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 (`"H"`). Thus,

`"H"=-16(3)^2+60(3)+6`

After some algebra...

`"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]}