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Finding maximum height of an object with no acceleration or velocity provided 
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#1
Sep1711, 09:27 PM

P: 5

1. The problem statement, all variables and given/known data
A football is kicked straight into up into the air; it hits the ground 5.2s later. What was the greatest height reached by the ball? Assume its kicked from ground level. With what speed did it leave the kicker's foot? 2. Relevant equations xf=xi + (vx)i + 1/2Ax(t)^2 3. The attempt at a solution I tried to solve by using the above formula, but had nothing to plug in for velocity, I know that g=9.8m/s, and I believe that that g would be the acceleration of the object, but I am unsure of how to progress in the problem. Any help would be greatly appreciated! 


#2
Sep1711, 09:40 PM

P: 400

You have two unknowns and one equation, so you need another equation. Hint: Energy is conserved.



#3
Sep1711, 09:46 PM

P: 329

You are correct that the acceleration is due only to gravity, so substitute g. You have nothing to substitute for v_{xi} because that is what you are looking for. Do you see that you know both x_{f} and x_{i}? With those you have enough information to solve. An alternate approach is to use the fact that the trip up and down are symmetrical and the velocity at the mid point is 0 m/s. If you can find the landing velocity, you also know the launch velocity. 


#4
Sep1711, 09:49 PM

P: 70

Finding maximum height of an object with no acceleration or velocity provided
We went over something similar to this in class this week. It looks to me like you'r eon the correct track with g. You know two of three parts of the v_{0}+at equation (with a being g), so find the unknown (v_{0}) and continue from there...



#5
Sep1711, 09:59 PM

P: 5




#6
Sep1711, 10:21 PM

P: 329

You had 0=0 + (V_{x})_{i}*(5.2s) 1/2g(t^{2}) From that it follows that 0=(V_{x})_{i}*(5.2s) 1/2(9.8 m/s^{2})(5.2s^{2}) Solving for (V_{x})_{i}, I get 25.48 m/s. 


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