Word problem

Question

Fountain shoots water from its base 25 ft in the air up to 50 ft into air. If the water takes 7 seconds to fall back to the ground from its highest point, write a function that models the water's height h at time t

Answer 1

h(t) = -1.0208(t - 4.95)^2 + 50

The vertex form of a quadratic function is

y = A(t- T) + k

 

Clearly k = 50 since this is the maximum height.

Let's find the acceleration.

It takes 7 seconds to fall from top to bottom. Initial velocity =0

t = 7, d = -50, vi = 0  a = ?

Using the equation of motion d = vit + 0.5 at^2

-50 = 0 + 0.5 x a x 7^2         100 = -49a      a = -2.04

 

Let's find the time that it takes the water to reach the maximum height. This is the same time that it takes the water to fall back from the maximum height to its starting position.

At the max. height, the speed is 0.  vi = 0    d = -25 feet,  acceleration = a = -2.04

Using the equation of motion, d = viT + 0.5 aT^2

-25  = 0 + 0.5 x (-2.04) xT^ 2           T = 4.95 seconds

 

The equation is therefore

y = A ( t- 4.95)^2 + 50

 

The total time of motion is t = 4.95 + 7 = 11.95 seconds. During this time the water reaches the ground.

We can use this time to find A.

Substitute into the equation above.

0 = A (11.95 - 4.95)^2 + 50            A = -1.02

y = h(t) = -1.02(t - 4.95)^2 + 50  

 

 

 

Answer 2

Lets break down the information using variables and logic.

 

 

At time t=0, the height h=25  (starting point)

 

It's maximum height is 50 feet.

 

At t=7, the height h=0.  (hits the ground)

 

 

Since we are looking for a quadratic function, the degree of this function will be 2.

 

Lets start with the maximum.  We know that to find the maximum, we need to find the vertex.  Start writing the function h(t) in vertex form.

 

h(t) = a(t - h)² + k

 

where:

a is the coefficient of the t² term.

Vertex has coordinate (h, k).

 

k is the y coordinate of the maximum.  c = h.

 

h(t) = a(t - c)² + 50

 

 

From here, we use the fact that  h(7) = 0.  So we expand the vertex form.

 

0 = a(7 - c)² + 50

 

0 = a(7 - c)(7 - c) + 50

 

0 = a(49 - 14c + c²) + 50

 

0 = 49a - 14ac + c² a + 50

 

 

Use the fact that h(0) = 25.

 

25 = a(0 - c)² + 50

 

25 = c² a + 50

 

 

We have just created two equations.

 

0 = 49a - 14ac + c² a + 50                    eq1

 

25 = c² a + 50                                        eq2

 

 

We can solve for c² a from eq2.

 

-25 = c² a

 

 

Substitute this value into eq1.

 

0 = 49a - 14ac - 25 + 50

 

0 = 49a - 14ac + 25

 

 

-25 = 49a - 14ac

 

 

Substitute eq2 into this equation to get only single terms.

 

-25 = 49(-25 / c²) - 14(-25 / c²)c

 

 

Solve for c from this equation.

 

-25 = (-1.96 / c²) + (350 / c)

 

-25c² / c² = (-1.96 + 350c) / c²

 

 

Equate numerators.

 

-25c² = 350c - 1.96

 

-25c² - 350c + 1.96 = 0

 

 

Solve for c using quadratic formula.

 

c = (350 ± √(122500 - (-49))) / -50

 

c = (350 ± 350.07) / -50

 

c = 0.0014           and           c = -14.0014

 

 

Plug in these values of c into eq1's vertex form to find a.  Keep in mind that a cannot be zero.

 

0 = a(7 - 0.0014)² + 50

 

-50 = 48.980a

 

-1.0208 = a

 

 

0 = a(7 + 14.0014)² + 50

 

-50 = 441.0589a

 

-0.1134 = a

 

 

The only equation of h(t) that has a starting height of  25 feet is

 

h(t) = -1.0208(t - 0.0014)² + 50