Conservation of Mechanical Energy - I can't see what I'm doing wrong

[Hemingway]

Homework Statement

A motorcyclist is trying to leap across a canyon. The first cliff is 70m high, she wants to jump to the cliff 35m high. She is traveling 38m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

I think (and am very worried by this) that my algebra is incorrect as I do not get the same answer as the textbook which is v = 46.2m/s

Homework Equations

1/2mv²+ mgh_f = 1/2mv₀²+mgh₀

The Attempt at a Solution

1/2mv²+ mgh_f = 1/2mv₀²+mgh₀

v²+ gh_f= u² + gh₀
v²+ (9.8) (35) = 38² + 9.8 (70)
v²= 1444 + 686 (-343)
v²= 1787
v = 42.3m/s

I know the equation is incorrectly reduced. The textbook has

(1) 1/2mv²+ mgh_f = 1/2mv₀²+mgh₀

to:

(2) vf = √(v₀² + 2g (h₀ - h_f)

I'm worried because it might mean I'm having a problem with fundementals - I would have just forgotten I used to be great at this stuff.

I have basically two questions:
- Is the final equation (2) not from (1) but just a form of the kinematics equation - so (1) cannot be reduced to (2)
- or what algebraic thing am I doing wrong? I cannot see how 2g happens as (1/2) / (1/2) = 1... I can't see what I can do to get 2g from equation 1.

Thanks as always to this forum!

 

[tsw99]

where is your factor of 2 in your 2nd step of part 3?

 

[Hemingway]

where is your factor of 2 in your 2nd step of part 3?

I'm not exactly sure with what you mean. How would I get a factor of 2? Perhaps there should be a factor of 2 but how do I get that? Once again, I am not sure how there is '2g' in the second equation...

:/

 

[tsw99]

I'm not exactly sure with what you mean. How would I get a factor of 2? Perhaps there should be a factor of 2 but how do I get that? Once again, I am not sure how there is '2g' in the second equation...

:/

1/2mv2+ mghf = 1/2mv02+mgh0

v2+ 2ghf= u2 + 2gh0

 

[Hemingway]

I'm not exactly sure with what you mean. How would I get a factor of 2? Perhaps there should be a factor of 2 but how do I get that? Once again, I am not sure how there is '2g' in the second equation...

:/

I suppose I need the working from equation (1) to it's manipulation to be equation (2) - as I just don't conceptually understand how the g's don't cancel if the mass cancels...

 

[Hemingway]

I'm not exactly sure with what you mean. How would I get a factor of 2? Perhaps there should be a factor of 2 but how do I get that? Once again, I am not sure how there is '2g' in the second equation...

:/

1/2mv2+ mghf = 1/2mv02+mgh0

v2+ 2ghf= u2 + 2gh0

Okay. Thank you for showing me that. I still don't understand why that is the case... I don't want you to waste your time showing me as you've already taken the time :) but is there a url I can look at to see why the '1/2's don't cancel but instead = 2 ?? I'm just a bit stuck I'm afraid :)

 

[tsw99]

Divide the whole equation by 1/2, m is a common factor so we cancel it, but g is not

I suggest you to improve your algebra first

 

[Hemingway]

Divide the whole equation by 1/2, m is a common factor so we cancel it, but g is not

I suggest you to improve your algebra first

Oh yes, that was never in dispute - I believe I prefaced the problem with that statement. I'll take a look. Thanks for the insight tsw99 :)

 

[Hemingway]

Oh yes, that was never in dispute - I believe I prefaced the problem with that statement. I'll take a look. Thanks for the insight tsw99 :)

okay I worked out how we get to...

v² + 2gh_f = u² +2gh₀
v² = u² + 2gh₀ - 2gh_f

Can anyone show me why +2gh doesn't cancel by showing me what does happen step by step? Alternatively can someone give me a url website/video that I can look at that will show me the princples as to how this equation ends up as that in my question above?

 

[Dadface]

Horizontally the bike moves with a constant 38m/s(ignoring air resistance).Vertically the bike gains speed due to gravitational attraction.You can find the speed gained by energy considerations (as tried above)or by using an equation of motion.Whatever method is used it must be remembered that the initial vertical velocity is zero(this is implied in the question).From this mg(H1-H2)=(mv^2)/2.From this v^2=2g*35.To find the speed on landing add the horizontal and vertical speeds vectorially by using Pythagoras.Is this the method you was using?

 

[Dadface]

I just looked more closely at your method and your mistake seems to be in line two of your attempt.To quote tsw 99 "where is your factor of 2?"

 

[Hemingway]

Oh yes, that was never in dispute - I believe I prefaced the problem with that statement. I'll take a look. Thanks for the insight tsw99 :)

I just looked more closely at your method and your mistake seems to be in line two of your attempt.To quote tsw 99 "where is your factor of 2?"

Thanks so much for your reply. As I said above, I am drawing an algebraic blank from how one gets from the first equation in my first post, to the second equation which is:
v² = u²+2g (h0-hf)

I know it's elementary, but I know how we get to v² +2gh_f =u²+ 2gh₀ but not to v² = u²+2g (h0-hf)

So I guess I am asking someone to humour me and show me how to get to the final equation. I know it's basic algebra and I'm revising as we speak and have been for the last few hours - but it's just my brain has dumped it in an irretrievable place at this stage...

So.. in conclusion - if someone could show step by step how I can manipulate the first equation to reach the final equation I would be most grateful.

 

[Hemingway]

basically I cannot see why 2g would not cancel ... and if I knew what I needed I'd recognise how absurd that statement is - but if someone could condescend to show me how step by step I'd be so, so appreciative. Sorry, just having one of those self-confidence draining moments :)

 

[Hemingway]

oh my gosh. I must have been in a haze of stupidity. I can't believe I couldn't do this. Fresh eyes makes a difference. Thanks to those hanging in there. Oh my - how embarrassing.

 

[Dadface]

You agree with tsw 99 that:
v2+2ghf=u2 +2gho.
To make v2 the subject of the equation get everything else to the other side of the equation by subtracting 2ghf from both sides.Now you have:
v2=u2+2gh0-2ghf.
Now you can put the numbers in and calculate it from there.Try it,it will give you the right answer.Now look again at the equation and you will see that 2g is common to to the last two terms on the right hand side so instead of writing 2gh0-2ghf you can write:
2g(ho-hf).This makes the final equation a little bit neater and saves a small step in the calculation.

 

[Dadface]

oh my gosh. I must have been in a haze of stupidity. I can't believe I couldn't do this. Fresh eyes makes a difference. Thanks to those hanging in there. Oh my - how embarrassing.

Look at the times of our two posts above Hemingway.You just beat me to it.