# Thread: Another mechanics question.

1. ## Another mechanics question.

In the figre the curve AB is a snow slope mounted on a frame used for ski jumping practice. Skiers start from rest at A, 30 metres above the horizontal ground and launch themseves into the air from B, 10 meters above the ground at 10 degrees. C is vertically below B and D is the point the skier will land if no air resistance/friction
Im guessing you use potential energy and kinetic energy to find V.
I then find V as 14m/s...
The answer is 35.5cm which im struggling to get.
The red part on the diagram is the projection.
Any help is appreciated.
Thanks

2. Originally Posted by AshleyT
Im guessing you use potential energy and kinetic energy to find V.
I then find V as 14m/s...
The answer is 35.5cm which im struggling to get.
The red part on the diagram is the projection.
Any help is appreciated.
Thanks
First we need to convert all units into meters .

then we have that $\displaystyle mg \Delta h =\frac{1}{2}mv^2 \iff v=\sqrt{2g \Delta h}=\sqrt{2g(.2)} \approx 1.97989$

This is the velocity at take off. Now the skier is a projectile

Now in the horizontal direction the only force is mass times gravity. Using the kinemetic equations to find when the skier hits the ground we get

$\displaystyle 0=.1+1.97989\sin(10^\circ)t-\frac{1}{2}(9.8)t^2$

Solving for t $\displaystyle t \approx .18218$

Now in the x direction there are no forces so we get

$\displaystyle x=1.97989\cos(10^\circ)(.18218) \approx 0.35522$

dont forget that the above answer is in meters and (unlike me) don't surpress your units. So we get $\displaystyle 35.5 cm$

3. Ah, thankyou very much. I was doing the right thing but wrong values =/...