Calculate Speed of Football at 5.0m Above Ground

[physical?eer]

a .565kg football is kicked from the ground at a speed of 22 m/s. what is its speed just as it passes between the uprights 5.0 m above the ground?

please help!

G:
m= .565 kg
v= 22m/s
5.0 m

r:
velocity

a:no kinematics's equations

 

[hage567]

What about conservation of energy? What do you know about that?

 

[physical?eer]

the only formulas i have are these

Energy Total = Energy kenetic + Energy Gravity

Ek = 1/2 mv^2

Eg = mgh

P=E+W

W= mgh

 

[hage567]

OK, the first of those are useful here. What do you think you should do with them? What does conservation of energy mean?

 

[physical?eer]

conservation:
energy can't be made or destroyed
total energy of any closed system remains same
energy can change from one from to another but total energy is same

 

[hage567]

So if the ball is on the ground, what does that tell you about the gravitational potential energy? Can you find the kinetic energy the ball has right after it is kicked?

 

[physical?eer]

well
Et= ek + eg
i know my v1 i think is 22m/s or is that my v2?

my h is 5.0
umm and my mass is .565

if i understood the velocity part that would help i think
cuz ek = 1.2 mv^2 but
v is v2-v1 and v2 is either 0 or 22

 

[physical?eer]

and no gravitational energy

 

[hage567]

well
Et= ek + eg
i know my v1 i think is 22m/s or is that my v2?

my h is 5.0
umm and my mass is .565

if i understood the velocity part that would help i think
cuz ek = 1.2 mv^2 but
v is v2-v1 and v2 is either 0 or 22

Your initial velocity (v1) is 22 m/s

Think of it in two parts. The first part is when the ball in on the ground right after being kicked. The second is when the ball is at the 5 m height. Write out the energy terms for each part.

So for first part Et1 = ek1 + eg1
for the second Et2 = ek2 + eg2

 

[hage567]

and no gravitational energy

For the first part, when the ball is at ground level, that is correct.

 

[physical?eer]

ET1 = 0
Et2 = 136.73 (1/2 * .565kg * 22m/s^2)
then...

 

[physical?eer]

wait hold on i think i missed it so at 5.0 m above ground its 22m/s right

 

[hage567]

Et1 is the sum of the two kinds of energy. When the ball is kicked, it has kinetic energy, and no potential energy since it is on the ground. So Et1 should be what you calculated for Et2.

Et2 will also be the sum of the kinetic and potential energies, but this time, the potential energy is NOT zero, since the ball is now 5 m above the ground. So you are given enough information to find the potential energy of the ball at this point, and you need to find the new kinetic energy.

 

[hage567]

wait hold on i think i missed it so at 5.0 m above ground its 22m/s right

NO. At 5 m above the ground, you don't know the velocity. That is what you are trying to find.

 

[physical?eer]

k i calculate Eg to be 27.685
then i looked at kinetic energy...but that leaves 2 variables velocity and kinetic energy in the formula

Ek = 1/2 mv^2

hmm...

 

[hage567]

You know that the initial kinetic energy of the ball must be equal to the sum of the kinetic and potential energy when the ball goes through the uprights. So you can find the final kinetic energy of the ball, since you know the potential energy. Once you have that, you can solve for the velocity. There is only one unknown here.

 

[physical?eer]

You know that the initial kinetic energy of the ball must be equal to the sum of the kinetic and potential energy when the ball goes through the uprights. So you can find the final kinetic energy of the ball, since you know the potential energy. Once you have that, you can solve for the velocity. There is only one unknown here.

sorry i don't understand... how does Ek1 = Ek + Eg and even then i don't no Ek

 

[hage567]

Because energy must be conserved. If the ball ONLY has kinetic energy to begin with, that means when it is above the ground at 5 m, some of that initial kinetic energy goes into gravitational potential energy, and some remains as kinetic energy. You are trying to find the FINAL kinetic energy, it will be different that the initial kinetic energy. Once you find the final kinetic energy, you can then find the velocity the ball has at 5 m above the ground.

 

[physical?eer]

so therefore Ek2 =
et1-eg
Et1 = 136.73 (1/2 * .565kg * 22m/s^2)

so ek2 =109.045 (136.73-27.685)
and so there fore...

 

[physical?eer]

i got 19.6468827 as velocity

 

[physical?eer]

sounds good...is it?

 

[hage567]

so therefore Ek2 =
et1-eg
Et1 = 136.73 (1/2 * .565kg * 22m/s^2)

so ek2 =109.045 (136.73-27.685)
and so there fore...

OK, so now you know the kinetic energy of the ball when it goes through the uprights. Now you can find the velocity, using the equation for kinetic energy (Ek = 0.5*m*v^2).

 

[hage567]

i got 19.6468827 as velocity

That looks reasonable. Don't forget to include your units in your answer.

 

[physical?eer]

was i wrong?

 

[physical?eer]

o i just saw your reply!
awesome
thank you soooo much!

 

[hage567]

was i wrong?

No, you just posted your answer before I saw it. I didn't realize you have calculated it already.

 

[hage567]

You're welcome.