Solve Energy/Work Problem: Find Height of Thrown Ball

[Tuffman087]

I am stumped by a problem I was given to find the height of a thrown ball from the point of release. The problem is:

During a contest that involved throwing a 7.0 kg bowling ball straight up in the air, one contestant exerted a force of 810 N on the ball. If the force was exerted through a distance of 2.0 M, how high did the ball go from point of release?

I've been working for the past 15 minutes trying to set up an equation/s, but I can't figure out which way to take the problem. Any advice on how to arrange the given quantities would be appreciated.

 

[bjr_jyd15]

Hmm...I'm not exactly an expert...lol. But I'll give you my advice anyway.

Draw a free body diagram on the ball. Find the sum of the forces (810 N going up, 700 N down)...remember W=F(distance)..

 

[bjr_jyd15]

Now, I'm pretty sure work=kinetic engergy. try to find the initial veloc. of the ball. then use kinematics to find max height.

 

[bjr_jyd15]

let me know if you understand what I'm saying...please show your work.

 

[Tuffman087]

I set it so that Work= Change in Kinetic Energy and got an answer I don't think is feasible. Fd=m(v squared)/2 (810)(2.0)=(7.0)(v squared)/2 v=square root(3240/7) which equals a velocity of 22 m/s after I plugged that into the Ke=Ug equation I got a height of 25 m.

I have a feelign I'm making this problem a lot harder than it truly is.

 

[bjr_jyd15]

remember the sum of the force is just 810-700 (weight of ball)...=110

 

[whozum]

At point of release KE = 1620J, PE = 0
At point of impact KE = 0, PE = 1620J
KE = .5mv^2
PE = mgh
PEi + KEi = PEf+KEf
mgh+.5mv^2 = mgh+.5mv^2. Notice both sides equal 1620.

mgh = 0 ______ .5mv^2 = 0

.5mv^2 = mgh

F*d = mv^2​

.5v^2 = gh

v^2 = F*d/m​

F*d/(2m) = gh
h = F*d/(2*m*g)
h = (810)*(2)/(2*7*9.8)
h = 1620/137.2
h = 11.8m

I hope this helps.

 

[compute_a_nerd]

Now, I'm pretty sure work=kinetic engergy. try to find the initial veloc. of the ball. then use kinematics to find max height.

Please correct me if I'm showing my stupidity here, but if the above statement is true, then would whozum's equation be incorrect in that
"F*D=mv^2" or would it be F*D=.5mv^2,
which would make it:

F*D/m=gh --> h=F*D/m*g

Thanks
Chris

 

[whozum]

For some reason when I wrote that, I had a reason that the 1/2 shouldn't be there, but now that I'm looking back at it, I have no idea why I omitted it.

You are correct.

[tex] KE = Fd = 1/2 mv^2 [/tex]

 

[compute_a_nerd]

thanks for clarifying that whozum