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?

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 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.

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.