Question

A ski jumper starts from rest from point A at the top of a hill that is a height h1 above point B at the bottom of the hill. The skier and skis have a combined mass of 80 kg. The skier slides down the hill and then up a ramp and is launched into the air at point C that is a height of 10 m above the ground. The skier reaches point C traveling at 42ms.Find his speed at the bottom of the hill which is 10m below the top.

Answer 1

Answer:

$v_i = 44.3 m/s$

Explanation:

As we know that here no friction force is present on the skier so we can say that total mechanical energy is conserved here

so we will have

$\frac{1}{2}mv_i^2 + mgh_1 = \frac{1}{2}mv_f^2 + mgh_2$

now we will have

$h_1 = 0$

$v_f = 42 m/s$

$h_2 = 10 m$

now we have

$v_i^2 = v_f^2 + 2gh_2$

$v_i^2 = 42^2 + 2(9.8)(10)$

$v_i = 44.3 m/s$

Answer 2

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The speed of the ski jumper at the bottom of the hill is "44.27 m/s".

The law of conservation of energy states that the energy always remains constant, although it may change from one form to another form. Applying this law between the point C and the bottom of the hill, we get:

Loss in Potential Energy = Gain in Kinetic Energy

$mg\Delta h = \frac{1}{2}m(v_f^2-v_i^2)\\\\2g\Delta h = v_f^2-v_i^2\\\\v_f^2=2g\Delta h+v_i^2\\\\v_f=\sqrt{2g\Delta h+v_i^2}$

where,

vf = speed at the bottom of the hill = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = loss in height = 10 m

vi = initial speed at the top of the hill = 42 m/s

Therefore,

$v_f = \sqrt{(2)(9.81\ m/s^2)(10\ m)+(42\ m/s)^2}$

vf = 44.27 m/s

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.