# Have a question about quadratic function (Algebra 1) Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with her center of...

Have a question about quadratic function (Algebra 1)

Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with her center of gravity 4 feet above the ground. She is thrown with an initial vertical velocity of 30 feet per second.

Questions

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?

What would the answer be?

If it's possible, I'd like to know how to solve the question.

Thank you

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### 3 Answers

We are given an object with projectile motion, directly up, with initial height 4 feet and initial velocity 30 feet per second.

The formula for height is:

`h(t)=-at^2+v_0t+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, etc...

(a) Will she reach 20 feet?

Let h(t)=20 and solve:

`20=-16t^2+30t+4 `

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

Using the quadratic formula we can solve for t:

`t=(-30 pm sqrt(30^2-4(-16)(-16)))/(2(-16)) `

**The discriminant is -124 so there are no real solutions. The girl will never reach 20 feet with the given initial velocity.**

The graph of height vs time:

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

Let h(t)=25 and solve for v0:

`25=-16t^2+v_0t+4 `

`-16t^2+v_0t=21 `

`v_0t=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 second.

With just the tools of Algebra I, you could graph the function and estimate the minimum:

or you could use a guess and revise strategy Just substitute values for v0 into `25=-16t^2+v_0t+4 ` until you find a solution.

**The best way using Algebra I is to use the discriminant.**

`-16t^2+v_0t-21=0 `

For there to be a real solution, the discriminant must be nonnegative.

`(v_0)^2-4(-16)(-21)>=0 ` then:

`(v_0)^2>=1344 ==> v_0>=sqrt(1344)~~36.66 `

Thus we need an initial velocity of about 37 feet per second.

**Sources:**

A quadratic function has the form:

y = **a**x^2 + **b**x + **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 posision of the body with respect to time, is described by a quadratic function of the form:

h = ±(1/2)gt^2 + V0t + h0

The ± signs correspond to the acceleration of gravity; 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** = V0 , **c** = h0

To solve the posed situation, we will transform the equation of motion, considering that the acceleration **g** is defined as the variation of the velocity divided between time:

g = (V – V0)/t → t = (V – V0)/g

Substituting **t** in the equation of motion, we have:

h = (g/2)[(V – V0)/g)]2 + V0(V – V0)/g + h0

After developing this equation, we get:

h = (V2 – V02)/2g + h0

Knowing that the final velocity of the object is zero to maximum height, and that the movement is upward, we substitute the given values:

a)

h = [(V2 – V02)/2g] + h0

h = [(–V02)/(-2g)] + h0 = (30)2/2(32.81) + 4

h = 17.72 feet

**So with an initial speed of 30 f/s, the object never reaches 20 feet**

b)

To find the initial velocity required to reach 25 feet, we will isolate **V0** in the above equation:

h = [(V2 – V02)/(-2g)] + h0

V02 = 2g (h - h0)

V0 = √[2g (h - h0)] = √[2(32.81)(25 – 4)]

V0 = 37.12 feet/s

**So to reach a height of 25 feet, must be launched with an initial speed of 37.12 ft/s**.

In my answer, I used the typical textbook simplification of the acceleration due to gravity -- using g=-32 ft per second squared. The actual value varies by location (depending on local topography as well as underlying densities), but the standard is 32.1740 feet per second squared.

You are using g=32.81 instead of 32.17. The conversion from cm per second squared to ft per second squared is .0328084. Is this where your value came from?

**Sources:**