Energy conservation equation to find equation for final velocity

[Racoon5]

Homework Statement

Use an energy conservation equation to find an expression for the skier’s speed as she flies off the ramp at point D. There is negligible friction between the skis and the ramp, and you can ignore the air resistance

Relevant Equations

KE = 1/2 mv²
PE = mgh
Energy conservation: E_A = E_D

I initially thought about the different forms of energy present at each of the points:
Total energy at starting point: PE_A+ KE_A= mgH

at point D:
KE_D = 1/2mv²_f PE_D= mgD
Energy at point D: PE_D+ KE_D
D = mgD + 1/2 mv²_f

because E_A= E_D

mgH = mgD = 1/2 mv²_f
mg(H-D) = 1/2 mv²_f
g(H-D) = 1/2 v²_f
so my V_f
came out to be the √2g(H-D)

I have checked this with one of my friends work and he has got a completely different answer to this. This seems logical to me, but I'm new to physics so I would appreciate someone pointing me in the right direction (not providing the solution). Thanks everyone!

 

[Orodruin]

You are correct.

 

[haruspex]

Energy conservation: E_A = E_D

There is not a point A. $E_H=E_D$?

mgH = mgD = 1/2 mv²_f

Presumably you meant mgH = mgD + 1/2 mv²_f

came out to be the √2g(H-D)

What you have written means (√2)g(H-D).
To show it correctly you can write √(2g(H-D)) or use LaTeX:
$\sqrt{2g(H-D)}$

 

[Racoon5]

There is not a point A. $E_H=E_D$?

Presumably you meant mgH = mgD + 1/2 mv²_f

What you have written means (√2)g(H-D).
To show it correctly you can write √(2g(H-D)) or use LaTeX:
$\sqrt{2g(H-D)}$

Thanks! You have a good point calling it point H. Perhaps I should define the points (A = initial, B = end of ramp, C = on the ground).

 

[haruspex]

Thanks! You have a good point calling it point H. Perhaps I should define the points (A = initial, B = end of ramp, C = on the ground).

Yes, that is better than using the same symbol for a point and its height.
If H and D had not been given, I would have used uppercase for the points (A, B say) and lowercase for the heights: a, b.