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A ball is launched vertically up from the ground, how long does it stay in the air? 
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#1
Nov512, 09:55 PM

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1. The problem statement, all variables and given/known data
A ball is launched vertically up from the ground. The initial speed of the ball is 37.5m/s. How long does it stay in the air? 2. Relevant equations The 5 kinematic equations 3. The attempt at a solution I figured the final speed would be 0m/s, and used the formula vf=vi+a*t so 0=37.5+(9.8)t and got t=3.83s... I don't think that's right, and was wondering if someone could please help me out? Any help is appreciated. Thanks 


#2
Nov512, 10:01 PM

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#3
Nov512, 10:06 PM

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#4
Nov512, 10:15 PM

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A ball is launched vertically up from the ground, how long does it stay in the air?



#5
Nov512, 10:20 PM

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#6
Nov512, 10:23 PM

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If you're stuck on a question, just always state the given even if you have no idea what to do! Given: a = 9.81ms_{2} (I prefer to write it like this on one line, one ms_{2} is equal to a m/s/s or a m/s^{2} v_{i} = 37.5ms_{1} d = ? v_{f} = 0 (at the top of its trajectory, the ball stops momentarily, then comes back down) v_{f}^{2} = v_{i}^{2} + 2ad Find the displacement, and then you can use a different equation to solve for time. (You now have four variables. Before, the three variables you had in your given restricted you to the use of the only equation that does not have time. Solving for displacement opens up the possibility to more equations, four of which include time) I had a feeling I just undermined what phantom said, so attempt the question again before you look at the spoiler please :)
Spoiler
Pick one of those equations and solve for t. Hint: I would pick d = v_{f}t  ^{1}/_{2}at^{2} this way, the final velocity of zero cancels out the vt term entirely, and you can solve without using the quadratic equation. 


#7
Nov512, 10:27 PM

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Try using one of the other kinematic equations, the one with displacement as a function of initial velocity, time and acceleration. The initial velocity and the acceleration of gravity are known, and what it is the value of the displacement between its initial position when it is thrown from the ground and its final position when it comes back down to the ground? Solve for t.....



#8
Nov512, 10:41 PM

P: 11

Hey guys, I found the displacement to be 71.75m. I used the equation suggested by Fifty and did this:
d = vft  1/2at2 71.75 = 0t  1/2(9.8)t2 71.75 = 4.9t2 I divided 4.92 from both sides so after I had this: 14.64 = t2 I took the square root of 14.64 and got 3.826s, rounded off to 3.83s What do you guy think? Did I make any mistakes? 


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Nov512, 11:06 PM

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#10
Nov512, 11:25 PM

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