Investigatin a ski jump

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Scott Jenkins

Investigating a ski-jump

Aim:To investigate how the starting position of a ski jumper affects the horizontal distance travelled in the jump. I will not take air resistance, friction and other various type of energy lost into account, however in practical I have to keep in mind that they do exist and cause variation in my results.

Introduction

Ski jumping is a sport event that involves a steep ramp and a landing zone, where the skier has to travel as far as possible after leaving the ramp horizontally. When the skier is in motion in the air and the range it reaches is what I am investigating. This motion is called the projectile motion and the displacement, velocity and acceleration of the projectile are all vector quantities. Each of these can be placed into vertical and horizontal components. In my experiment, I will create a similar model of the ski jump using a plastic curtain rail as the slope and model the skier as a particle, in this case, a ball bearing.

Diagram                                    This is a diagram of the basic equipment setup

This investigation has much room for expansion on the original above setup. The accuracy can be improved using a combination of more sensitive measuring equipment and a more accurate measuring setup.

Background Knowledge

Speed = Distance                                        Re-arrange                    

                  Time                                                                            

Range = Velocity x Time

X= VT……….Equation 1

To find the range (x), the velocity and time must be found.

m=mass, kg
h=height1, m
H=height2, m
v=velocity, ms
-¹
s=displacement (the range), m
u=initial velocity, ms
-¹
t=time, s
a=g=9.8, ms
-²

Assuming no energy is lost, potential energy is equal to kinetic energy by the law of conservation of energy.

Velocity

GPE=KE                        (GPE= mgΔh)
mgΔh=½mv²                        (KE=½mv²)
mgΔh=½mv²                        Masses cancel
½v²=gΔh                        x2
v²=2gh                                    √

v=√(2gh) ……….. Equation 2

Time

By using the equation of linear motion, s=ut+½at²

s=0+½at²                        vertical velocity = 0
H=½at²                        a=g
H=½gt²                        x2
2H=gt²                                     ÷ g
2H/g=t²                                    √

t=√ (2H/g) ………..
Equation 3

Join now!

Equations 2 and 3 can now be substituted into equation 1.

Range(x)=Velocity(v) x Time(t)
R=√(2gh) x √(2H /g)                                                  g cancels out
R=√(2h) x √(2H)

R=√(4hH)…………
Equation 4

Assumptions

  1. The friction between the ball and the curtain rail, along with the air resistance are neglible, therefore, will not be accounted for in this experiment.
  2. The ball bearing is taken as a point, as the motion of a rolling ball differs from anything else.
  3. Gravity does not affect the horizontal velocity

Prediction/Hypothesis

I have based my ...

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