Forces and Motion

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  • Topic: Olympic Games, International Olympic Committee, Pierre de Coubertin
  • Pages : 5 (1567 words )
  • Download(s) : 166
  • Published : May 25, 2013
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Shot put is a track and field event involving „throwing“/“putting“(throwing in a pushing motion) a heavy spherical object (the shot) as far as possible.1 The shot put has ancient origins in strong man competitions, in which stones were originally thrown instead of metal balls.

We have found that force of gravity, and acceleration play a role to a flying object. Force of gravity affects the object vertically while force of acceleration may differ between how the object is being thrown. In fact, it will always change unlike force of gravity. These forces affect everything we do, from walking to class to passing a pencil to a friend. In relation to everyday life, there is a clear connection to many kinds of sports. Projectile motion can be applied to any object due to the forces the athletes apply on it. By knowing these forces, it gives the athlete an advantage to know where the object might be later on. For example, from a athlete’s perspective, acquiring experience from becoming more intuitive with an objects motion will aid him or her in knowing how far the object might travel. The athlete will know how much force to apply on the object to score a goal. Of course the athlete cannot calculate the exact amount of force during a game so the athlete will have to use instinct and previous experience to estimate how much force to apply to the object.

Shot put field

CALCULATIONS
For the case of this paper I will assume that:
* athlete can produce always the same shot put force (F)
* air friction does not exist and equals zero, I will neglect air resistance Shot put throw is made by to major forces that accelerate ball, vertical and horizontal and video of such throw is here http://youtu.be/F6vHMDXjKyQ. Trajectory of the shot put ball is equal to trajectory of shooting a canon

I will show that successful athlete is one than can not only throw ball with greatest force but one that can also control angle of the shot, not to horizontal, not to vertical. Athlete throws ball at angle (Θ) or we can say athlete gives ball initial velocity v0. This velocity can be broken into vectors, horizontal and vertical. Both velocity vectors are results of forces that athlete gave ball during the shot, vertical and horizontal vector force, F=Fv+Fh. To better understand shot put we have to break the initial force (or velocity) into horizontal (x) and vertical (y) components: Horizontal component is Fx=FcosΘ and vertical component is Fy=FsinΘ and same analogy can be used on initial velocity (v0): Horizontal component is vxo= v0cosΘ and vertical component is vyo= v0sinΘ. When throw is made gravity force (G) starts to act on ball and causes ball to fall, to stop. Gravity force is vertical and has influence only on given vertical force component. Vertical initial force (Fy) accelerated ball to initial vertical velocity (vy0). Firstly gravity will retard ball to stop and then ball will start to accelerate in opposite vertical direction. When vertical velocity stops (vy=0) ball will be in the highest point. Therefore we can conclude that hang time (t) in air consist of two parts: ball retardation and ball acceleration. We must not forget that height of retardation does not equal height of acceleration hence ball is thrown from the shoulder (ie 160cm) and therefor length of fall will be longer depending of the height of athlete shoulder. Horizontal velocity of the ball will not change because we assumed that no forces will retard this component. Horizontal velocity is uniform velocity.

In respect to stated I will calculate distance of the throw assuming that initial velocity and the angle of the throw are known. I will calculate distance using formula s=vx0t (where t represents hang time).

Hang time is sum of retardation and acceleration t=tr+ta
Horizontal component is vxo= v0cosΘ and vertical component is vyo= v0sinΘ. Retardation time is calculated by formula v= vy0-at, in this case 0= vy0-gtr where g is gravitational...
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