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 hatelove Feb22-12 05:54 PM

Solution to quadratic equation doesn't look right

1. The problem statement, all variables and given/known data

Quote:
 A pitcher throws a ball towards a wall that's 85 feet away from him. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the wall?
2. Relevant equations

distance = rate * time

3. The attempt at a solution

So the speed of the ball I have as:

[t = time]

5 + 32t

Which I believe makes sense because the ball is initially traveling @ 5 ft/sec, and after 1 second, the ball is traveling 37 ft/sec, then after 2 seconds the ball is traveling @ 69 ft.sec, etc.

I assume we use d = rt (which is distance = rate * time).

So the total distance the ball travels is 85 feet which is equal to the total rate of 5 + 32t * time. The equation I set up like this:

85 = (5 + 32t) * t

I set up the quadratic like so:

32t^2 + 5t - 85 = 0

I solved it and got:

(1/64) * (sqrt(10905) - 5) which is approx. 1.55.

I worked it out to check, and 1.55 seconds is much too low for the value:

5 feet + (32 feet * 2 seconds) = 69 feet

So after 2 seconds, the ball has only traveled 69 feet, while the quadratic equation states that the ball has traveled 85 feet after 1.55 seconds.

What gives?

 eumyang Feb22-12 06:09 PM

Re: Solution to quadratic equation doesn't look right

Quote:
 Quote by hatelove (Post 3778769) So the speed of the ball I have as: [t = time] 5 + 32t
Quote:
 Quote by hatelove (Post 3778769) 5 feet + (32 feet * 2 seconds) = 69 feet So after 2 seconds, the ball has only traveled 69 feet, while the quadratic equation states that the ball has traveled 85 feet after 1.55 seconds. What gives?
You're forgetting that 5+32t represents speed, not distance.

 Mark44 Feb22-12 08:17 PM

Re: Solution to quadratic equation doesn't look right

Quote:
 Quote by hatelove A pitcher throws a ball towards a wall that's 85 feet away from him. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the wall?
This sounds like a made-up problem that doesn't have much connection to reality. After the ball leaves the pitcher's hand, the only acceleration on it is due to gravity, which acts straight down. Maybe the ball happens to have a propellor or rocket motor attached? Is this the exact wording of the problem?

 HallsofIvy Feb23-12 08:55 AM

Re: Solution to quadratic equation doesn't look right

Because this is a two dimensional problem. you should break it into horizontal and vertical components. As Mark44 said there is an acceleration of -9.81 m/s^2 in the vertical direction and no acceleration in the horizontal direction.

 Char. Limit Feb23-12 12:20 PM

Re: Solution to quadratic equation doesn't look right

Quote:
 Quote by HallsofIvy (Post 3779915) Because this is a two dimensional problem. you should break it into horizontal and vertical components. As Mark44 said there is an acceleration of -9.81 m/s^2 in the vertical direction and no acceleration in the horizontal direction.
Or since he's using feet and seconds, an acceleration of about -32 ft/s^2 vertical and 0 horizontal.

 Mark44 Feb23-12 01:22 PM

Re: Solution to quadratic equation doesn't look right

Quote:
 Quote by hatelove A pitcher throws a ball towards a wall that's 85 feet away from him. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the wall?
As I said before, this is really a flaky problem, with almost no grounding in reality. I still wonder if the OP is showing the correct statement for this problem.

Has anyone noticed that the ball's speed is 5 ft/sec? That works out to about 3.4 mi/hr. If I start walking toward the wall at the same time the ball is thrown, I'll get there first (I can walk faster than 3.4 mph).

Furthermore, I can't imagine any scenario in which the ball would actually get to the wall, inasmuch as it will fall ~4600 ft during its flight.

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