Law of conservation of energy homework help

[hasek888]

Law of conservation of energy homework help!

Homework Statement

A skier goes down a slope and goes over a ramp. From a resting position which has a height of 16m, she leaves a ramp of 9m at an angle of 45 degrees and passes over a haystack forming a 30degree angle vertically. Supposing there is no friction and that the skier's dimension are negotiable, find the height of the haystack.

I attached a drawing of the problem also.

Homework Equations

ET1 = ET2
mgh1 + m(v1^2)/2 = mgh2 + m(v2^2)/2
I think that's what I am supposed to do...

The Attempt at a Solution

Well, so far I've done the following:
h1 = 16m
h2 = 9m
v1 = 0m/s
using the equation above,
9.8 x 16 + 0 = 9.8 x 9 - V2^2
V2 = 8.3 m/s
It's all i have now, and I am not even sure whether what I did is right or not!

Any help is greatly appreciated!

 

[Delphi51]

Welcome to PF, hasik.
I'm not quite following your calc. I'm thinking potential energy is converted to kinetic as the skier falls 7 m.
mgh = ½mv²
v = squareRoot(2gh) where h = 7
for the speed of the skier leaving the ramp.

The rest of the question isn't entirely clear to me!
In the diagram, it looks like we must find the height where the trajectory is 30 degrees from vertical. But the trajectory angle will then get closer to vertical as it goes further and hit the side of the haystack, contrary to the words in the question ("passes over a haystack"). What do you think - does that haystack "form a 30 degree angle" or is it shaped like a post as shown in the diagram?

Anyway, the next step is to write some equations for the trajectory of the flying skier. I like to write one horizontal equation
x = Vix*t (find the Vix from the initial speed and angle)
and two vertical equations
y = Viy*t + ½at² and Vy = Viy + at
Note that the angle at time t will be related to the Vx and the Vy, not the x and y.

 

[hasek888]

the haystack is a post. The 30 degree angle is formed by the skier.

what does x and y represent? and how am I supposed to find t?

 

[Delphi51]

Don't try to find t yet - you want equations that tell you what the velocities and position of the skier are at all times. Late you can solve for the time when the trajectory has the 30 degree angle.

Just put your known numbers into the three equations for starters.

 

[hasek888]

well, as the skier gets of the ramp, V1x and V1y are both equal to 8.3m/s.

1) x = 8.3t
2) y = 8.3t + 4.9t^2
3) V2y = 8.3 + 9.8t

Is this correct? If so, what are the next steps I should take?

 

[Delphi51]

Make g negative in 2) and 3).
You want the time when the velocity vector is at 30 degrees to vertical.
Sketch Vx and Vy vectors on a triangle, put in the 30 degrees and you'll get a trig ratio relating Vx and Vy. Solve that for t.

 

[hasek888]

You only have the angle, which is 30.
How can you possibly calculate Vx and Vy??

 

[Delphi51]

tan(30) = Vx/Vy
Put in the equations you have for Vx and Vy.
Solve for t.
With the known time, you can find the x or y position you need!

 

[hasek888]

So :
tan(30) = 8.3/Vy
Vy = 8.3/tan(30)
Vy= 14.4m/s

V2y = 8.3 -9.8t
14.4 = 8.3 - 9.8t
t= 0.622s

y = 8.3t - 4.9t^2
y = 8.3(0.622) - 4.9(0.622)^2
y = 3.26m

Therefore the haystack is 3.26m high??

 

[Delphi51]

Looks good!
Careful with that 3 digit answer - we were careless rounding things to 2 digits in the middle when we should have kept 4 digits if you need 3 at the end. And g = 9.81 unless your class convention is 9.8. The local college here uses 9.83, the actual local gravitational constant. It makes the answers come out slightly different from the ones in the back of the book.