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Finding maximum height of an object with no acceleration or velocity provided

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BioMan789
#1
Sep17-11, 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!
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JeffKoch
#2
Sep17-11, 09:40 PM
P: 400
You have two unknowns and one equation, so you need another equation. Hint: Energy is conserved.
Fewmet
#3
Sep17-11, 09:46 PM
P: 329
Quote Quote by BioMan789 View Post
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!
Welcome to Physics Forums, BioMan789.

You are correct that the acceleration is due only to gravity, so substitute -g. You have nothing to substitute for vxi because that is what you are looking for.

Do you see that you know both xf and xi? 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.

ohms law
#4
Sep17-11, 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 v0+at equation (with a being -g), so find the unknown (v0) and continue from there...
BioMan789
#5
Sep17-11, 09:59 PM
P: 5
Quote Quote by Fewmet View Post
Welcome to Physics Forums, BioMan789.

You are correct that the acceleration is due only to gravity, so substitute -g. You have nothing to substitute for vxi because that is what you are looking for.

Do you see that you know both xf and xi? 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.
When I try to solve for the velocity (Vx)i, I plug in 0 for xf (when the ball lands) and 0 for xi (assume the ball is kicked from ground level), and am left with 0=0 + (Vx)i*(5.2s) -1/2g(t^2) , then solving for x, I factored it to have -x(4.9x - 5.2)=0, and solved x, which is (Vx)i to be 1.06m/s. Apparently this is not the right answer (according to the provided answer by the physics question site). I don't know what I'm doing wrong.
Fewmet
#6
Sep17-11, 10:21 PM
P: 329
Quote Quote by BioMan789 View Post
When I try to solve for the velocity (Vx)i, I plug in 0 for xf (when the ball lands) and 0 for xi (assume the ball is kicked from ground level), and am left with 0=0 + (Vx)i*(5.2s) -1/2g(t^2) , then solving for x, I factored it to have -x(4.9x - 5.2)=0, and solved x, which is (Vx)i to be 1.06m/s. Apparently this is not the right answer (according to the provided answer by the physics question site). I don't know what I'm doing wrong.
There is some algebra problem there that I cannot figure out. Could it be that you are treating the V and x in Vx as two variables? The symbol (Vx)i is the initial velocity in the x-direction.

You had 0=0 + (Vx)i*(5.2s) -1/2g(t2)

From that it follows that
0=(Vx)i*(5.2s) -1/2(9.8 m/s2)(5.2s2)

Solving for (Vx)i, I get 25.48 m/s.


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