Can someone tell me how to solve this:

A particle is projected vertically upward from the earth' s surface with initial speed Vo. Prove that the maximum height H reached above the earth's surface is
H=Vo^2R( 2gR-Vo^2)

I think it can be solved using conservation of energy but I dont know where to start from. Please help!!!

Derivation of the Maximum Height Reached by a Projectile






Tbottom = mgH1 + 1/2 mV02 Energy is Conserved
Ttop = mgH2 + 1/2 m(Vtop)2
mgH1 + 1/2 mV02 = mgH2 + 1/2 m(Vtop)2 H1 = 0
1/2 mV02 = mgH + 1/2 m(Vcosq )2 Divide by m; m cancels
1/2 V02 = gH + 1/2 (Vcosq )2
V2 = 2gH + (Vcosq )2 Multiply both sides by 2
1/(cos2q ) = (2gH)/ (Vcosq )2 + 1 Divide both sides by
V2 = 2gH + (Vcosq )2 (Vcosq )2
(V2(1 - cos2q) / (2g) = H Multiply both sides by (Vcosq )2
Energy is Conserved
Ttop = mgH2 + 1/2 m(Vtop)2
mgH1 + 1/2 mV02 = mgH2 + 1/2 m(Vtop)2 H1 = 0
1/2 mV02 = mgH + 1/2 m(Vcosq )2 Divide by m; m cancels
1/2 V02 = gH + 1/2 (Vcosq )2
V2 = 2gH + (Vcosq )2 Multiply both sides by 2
1/(cos2q ) = (2gH)/ (Vcosq )2 + 1 Divide both sides by
V2 = 2gH + (Vcosq )2 (Vcosq )2
(V2(1 - cos2q) / (2g) = H Multiply both sides by (Vcosq )2

H = (V2(1 - cos2q) / (2g)

We can work out the maximum height in two ways. If we have been given, or just worked out, the time of flight, as in the previous section, we know that at exactly half this time it will be at the top of the trajectory. If we didn't have this information we could use the fact that the vertical velocity will be zero at the max height and use the formula for velocity to find the correct value for t.
Vy=Uy-gt

I am sorry but still confused. We didn't get H=Vo^2R( 2gR-Vo^2)
did we?

R is the earth's radius. If the gravitational attraction at the earth's surface is mg, then the attraction at some height r above the surface will be mgR^2/(R+r)^2. The attraction can be integrated from r = 0 to r = H to get the total work involved, which must equal the kinetic energy mVo^2/2. However, when we integral from
r=0 to r=H of mgR^2/(R+r)^2 and set it equal to mVo^2/2,
H= does not give the value Vo^2R/(2gR-Vo^2) ??

Hey pat, the process you've described for the solution is correct and I just solved it to get R(Vo)^2/(2gR-Vo^2)

After integrating, (P.E)at H = mgR^2 [ (1/R) - (1/(R+H)) ] <--- Did you get this?

Equating (P.E)at H with (K.E.)initial, you get the desired answer. I think you did the integral wrong, or did some algebra mistake while equating.

Possible mistake; integral(1/(R+r)^2, r) <--- did you change variable here and forget to change the limits of integration.

Really propensity, u got the answer....I got the same solution for the integrall....I think I screwed up in the algebra but still not getting the answer..please help...

ok,

mgR^2 [ (1/R) - (1/(R+H)) ]= 1/2 mv^2

or, gR^2[ (1/R) - (1/(R+H)) ]= 1/2 v^2

or, gR^2[ ((R+H)-R)/(R(R+H)) ] = 1/2 v^2

or, gR^2[ H/(R(R+H))= 1/2 v^2
..........
when you cross multiply it get all jumbled up and got me something else but not the required answer, can you please see where does it go now...............
Thanks

The instantaneous change in velocity at the top is zero. So, differentiate the distance equation to get the equation for velocity and equate it with zero.

do I make sense out here?

But there is no distance equation..

OK, When you are solving for H, don't make the problem more complex by increasing the number of H.

In your first and second equations of the last post, there are just one H, but you made the problem more complex by increasing the number of H in equation 3. Don't do that. Instead,

(1/R) - 1/(R+H) = (V^2)/(2gR^2)
or, 1/(R+H) = (1/R) - (V^2)/(2gR^2)
or, 1/(R+H) = (2gR - V^2) / (2gR^2)
or, H = (2gR^2)/(2gR-V^2) - R
or, H = (RV^2)/(2gR - V^2)

Awesome...Thank you so much prospensity....................

lots of physics majors

Oh man!!!
I think this was in high school right? Is this a college level question? I don't know, may be physics 101?

I ain't a physics major by the way. I hate science classes!!!

Now it University Physics 211. Been a long time, i remember looking for it in a solution guide for take home quiz.