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Some physics homework help plz

F

Finns14

Golden Member
Ok so here is the problem

A 51.3-g golf ball is driven from the tee with an initial speed of 52.9 m/s and rises to a height of 34.7 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 5.57 m below its highest point?

Ok well I know Ke= 1/2mv^2 but I still can't seem to figure out the answer. I am guessing that either 34.7 isn't its highest point of that I have to subtract Ke inital from Ke final and thats throwing me way off. I may be completely off all together. Any help would be greatly appreciated. Thanks
 
Well its an online homework thing so I can give you the answer but then it generates a new set a values so would the answer be helpful?
 
Originally posted by: Finns14
Well its an online homework thing so I can give you the answer but then it generates a new set a values so would the answer be helpful?
Click to expand...

If you have the final answer to the first set of values, you can work backwards. This will allow you to generate your own answer for the new set of values.
 
Originally posted by: BigJ
Originally posted by: Finns14
Well its an online homework thing so I can give you the answer but then it generates a new set a values so would the answer be helpful?
Click to expand...

If you have the final answer to the first set of values, you can work backwards. This will allow you to generate your own answer for the new set of values.
Click to expand...



tried that to no avail
 
Originally posted by: Finns14
Originally posted by: BigJ
Originally posted by: Finns14
Well its an online homework thing so I can give you the answer but then it generates a new set a values so would the answer be helpful?
Click to expand...

If you have the final answer to the first set of values, you can work backwards. This will allow you to generate your own answer for the new set of values.
Click to expand...



tried that to no avail
Click to expand...

Well what was the problem you originally tried?
 
a)


Number
54.3341385

Units
J

(b)


Number
47.195995592847

Units
m/s



ok thats the answer for the problem at the top of the thread
 
Vi = Initial Velocity

Total Energy = 1/2 mVi^2 + mgh

The total energy is the intitial kinetic energy because theball is on the ground.

V = sqrt(0.5mVi^2 - mgh)/(0.5m)
 
Originally posted by: jstack
Vi = Initial Velocity

Total Energy = 1/2 mVi^ + mgh

The total energy is the intitial kinetic energy because theball is on the ground.

V = sqrt(0.5mVi^2 - mgh)/(0.5m)
Click to expand...

Technically, the ball is not on the ground. It is on a tee at the start, so it has some potential energy at that point.
 
Originally posted by: BigJ
Originally posted by: Gibson486
at the highest point, velocity is 0, isn't it?
Click to expand...

Velocity in the Y direction, yes. Velocity in the X direction, no.
Click to expand...

isn't it just the Y direction we only care about? After all, the Velocity in the X axis will not change.....
 
Vf^2 = Vi^2 + 2ad (I think that's the right equation, going off memory)

So 0 = Vi^2 + 2*-9.81*34.7
Vi = 26.1 m/s (That being the initial vertical velocity)

So V^2 = Vx^2 + Vy^2
52.9^2 = Vx^2 + 26.1^2
Vx = 46.0 m/s

At the highest point, Vy = 0 so

Ke = 1/2*m*Vx^2
Ke = 1/2 * 0.0513 * 46^2

Ke = 54.3 J

Done!
 
Bahahah, I forgot that M is in KG.

54334 / 1000 = 54.334Js

The way to solve it using the PE and KE equations is simply subtracting the PE at the highest point from the KE final.

KE final = 1/2 * m * v^2
KE final = 1/2 * 51.3/1000 * 52.9^2
PE final = 0

Using the Conservation of Energy theorem

PE final + KE Final = KE point q + PE point q
0 + 1/2 * 51.3/1000 * 52.9^2 = KE point q + 51.3/1000 * 34.7 * 9.8
71.7792165 = KE point q + 17.445078
KE point q = 71.7792165 - 17.445078
KE point q = 54.3341385

You can do it using the final velocity formula in conjunction with KE formula, but if a professor is testing a concept, he may not accept it as an answer.
 
KE + PE = constant

KE + mgh (h= height in meters) = total energy

so when the ball is at max height, MGH = 1/2 MV^2

when the ball is below max height, MG(h-5.57) + 1/2MV^2 = original 1/2 MV^2

solve the problem yourself
 
Originally posted by: Fenixgoon
KE + PE = constant

KE + mgh (h= height in meters) = total energy

so when the ball is at max height, MGH = 1/2 MV^2

when the ball is below max height, MG(h-5.57) + 1/2MV^2 = original 1/2 MV^2

solve the problem yourself
Click to expand...

Why would PE = KE at max height? If that was the case, the answer would simply be KE final /2.
 
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