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Solution to quadratic equation doesn't look right 
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#1
Feb2212, 05:54 PM

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


#2
Feb2212, 06:09 PM

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#3
Feb2212, 08:17 PM

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#4
Feb2312, 08:55 AM

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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.



#5
Feb2312, 12:20 PM

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#6
Feb2312, 01:22 PM

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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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