The Math of Stunting

The physical nature of the sport cheerleading, demands a grasp on the knowledge of how both the forces that are exerted 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 pyramid. The highly organized nature of cheerleading, with all of its formations and different groupings of, in Columbia River’s case, members based on their “job” in that stunt – back spot, base, or flyer – can hardly be removed from the concepts of math. Rather, a cheerleader with a true knowledge of physics and mathematics will undoubtedly 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…*

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

-16t^2+30t-16=0

Using the quadratic formula, x=(-b±√(b^2-4ac))/2a , we can solve for t: x=(-30±√(〖30〗^2-4(-16)(-16)))/(2(-16)) The discriminant, x=-b±√(b^2-4ac) , is -124 so there are no real solutions. The flyer will never reach 20 feet with the given initial velocity.

The graph of height vs time:

Height (in feet)

Time (in seconds) (b) What initial velocity is required to achieve a height of 25 ft.?

Let h(t)=25 and solve for V_0: 25=-16t^2+v_0 t+4 -16t^2+v_0 t=21 v_0 t=21+16t^2 v_0=21/t+16t

We want to minimize this function. Using calculus we find the minimum to be at t approximately 1.15 seconds and velocity approximately 36.66 feet per

The physical nature of the sport cheerleading, demands a grasp on the knowledge of how both the forces that are exerted 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 pyramid. The highly organized nature of cheerleading, with all of its formations and different groupings of, in Columbia River’s case, members based on their “job” in that stunt – back spot, base, or flyer – can hardly be removed from the concepts of math. Rather, a cheerleader with a true knowledge of physics and mathematics will undoubtedly be more successful in their happenings –

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 =

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

-16t^2+30t-16=0

Using the quadratic formula, x=(-b±√(b^2-4ac))/2a , we can solve for t: x=(-30±√(〖30〗^2-4(-16)(-16)))/(2(-16)) The discriminant, x=-b±√(b^2-4ac) , is -124 so there are no real solutions. The flyer will never reach 20 feet with the given initial velocity.

The graph of height vs time:

Height (in feet)

Time (in seconds) (b) What initial velocity is required to achieve a height of 25 ft.?

Let h(t)=25 and solve for V_0: 25=-16t^2+v_0 t+4 -16t^2+v_0 t=21 v_0 t=21+16t^2 v_0=21/t+16t

We want to minimize this function. Using calculus we find the minimum to be at t approximately 1.15 seconds and velocity approximately 36.66 feet per

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