Question

The crest of the second hill is circular, with a radius of r = 35.5 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

Answer 1

When the skier loses contact on the second hill,the N(normal reaction due to hill) vanishes.Consider the forces acting on skier at the top of the hill.

1]Mg acts downwards
2]N upwards
Therefore,

\[Mg = m \frac{V^2}{r}\]
There is no N in the above equation because it vanishes when the skier leaves the contact
Therefore:
\[Vf = \sqrt{g*r}\]
Now apply Energy conservation at 2 pts at the top of two hills:
\[mgH = mg(35.5) + \frac{1}{2}m(Vf)^2\]

makes sense?

Answer 2

Solving for H at the end

Answer 3

u are the real man :)

Answer 4

thanks hehe

Answer 5

do they ask what the mass is ?

Answer 6

cause it's not needed, the mass cancels out oin the equation

Answer 7

ok