The Concept Of Stunting In Cheerleading
The Math of Stunting
The vigorous nature of the sport cheerleading, demands a hold on the understanding of how both the forces that are applied and those one experiences will affect one’s performance in a stunt or trick. Physics and mathematics can be found everywhere in cheerleading: 8-counts direct a dancing routine, weight distribution determines a three level stunt. The highly structured nature of cheerleading, with all of its formations and different groupings of, in Columbia River’s case, participants based on their job in that stunt – back spot, base, or flyer – can hardly be uninvolved from the concepts of math. Rather, a cheerleader with an accurate knowledge of physics and mathematics will undeniably be more successful in their happenings …show more content…
The flyer begins this move with their center of gravity 4 feet above the ground. One is thrown with an initial vertical velocity of 30 feet per second.
1. Will the flyer’s center of gravity ever reach 20 feet?
2. For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be?
A quadratic function has the form: y = 〖ax〗^2+ bx + c
The movement that makes this object is a uniformly varied movement, because it is subject to the acceleration of gravity. In these types of movement, the position of the body with respect to time, is described by a quadratic function of the form: h = ± (1/2) 〖gt〗^2+ V_0 t+ h_0
The ± signs correspond to the acceleration of gravity and the sign is chosen according to the direction of motion; if the motion is upward the value (-g) is taken and if it is down, the value (+g) is taken.
Comparing these equations, we see that both have the same structure:
The variable y represents the height h, and the variable x represents time t; also the coefficients a, b and c have the values: a = ± (1/2) g, b = V_0 , c = …show more content…
The formula for height is: h(t)=-at^2+v_0 t+s_0
Here a=16 (this is the gravitational force since we are measuring in feet a=16. If we measured in meters then a=4.9.) t is the time in seconds, v0 is the initial velocity of 30 feet per second, and s0 is the initial height or 4 feet. Thus we have: h(t)=-16t^2+30t+4 as a model for the height.
Note that we disregard complicating factors such as air resistance, spin, because the uncertainty and unpredictability of these factors leads to a much more complicated equation than needed.
(a) Will the flyer reach 20 feet?
Let h(t)=20 and solve:
20=-16t^2+30t+4
The vigorous nature of the sport cheerleading, demands a hold on the understanding of how both the forces that are applied and those one experiences will affect one’s performance in a stunt or trick. Physics and mathematics can be found everywhere in cheerleading: 8-counts direct a dancing routine, weight distribution determines a three level stunt. The highly structured nature of cheerleading, with all of its formations and different groupings of, in Columbia River’s case, participants based on their job in that stunt – back spot, base, or flyer – can hardly be uninvolved from the concepts of math. Rather, a cheerleader with an accurate knowledge of physics and mathematics will undeniably be more successful in their happenings …show more content…
The flyer begins this move with their center of gravity 4 feet above the ground. One is thrown with an initial vertical velocity of 30 feet per second.
1. Will the flyer’s center of gravity ever reach 20 feet?
2. For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be?
A quadratic function has the form: y = 〖ax〗^2+ bx + c
The movement that makes this object is a uniformly varied movement, because it is subject to the acceleration of gravity. In these types of movement, the position of the body with respect to time, is described by a quadratic function of the form: h = ± (1/2) 〖gt〗^2+ V_0 t+ h_0
The ± signs correspond to the acceleration of gravity and the sign is chosen according to the direction of motion; if the motion is upward the value (-g) is taken and if it is down, the value (+g) is taken.
Comparing these equations, we see that both have the same structure:
The variable y represents the height h, and the variable x represents time t; also the coefficients a, b and c have the values: a = ± (1/2) g, b = V_0 , c = …show more content…
The formula for height is: h(t)=-at^2+v_0 t+s_0
Here a=16 (this is the gravitational force since we are measuring in feet a=16. If we measured in meters then a=4.9.) t is the time in seconds, v0 is the initial velocity of 30 feet per second, and s0 is the initial height or 4 feet. Thus we have: h(t)=-16t^2+30t+4 as a model for the height.
Note that we disregard complicating factors such as air resistance, spin, because the uncertainty and unpredictability of these factors leads to a much more complicated equation than needed.
(a) Will the flyer reach 20 feet?
Let h(t)=20 and solve:
20=-16t^2+30t+4