How do I create a function and graph on a coordinate plane? I'm hopeless.

Question

A water ballon is catapulted into the air so that its height h, in meters, after t seconds. Using the acceleration due to gravity, the initial velocity of 27 meters, and an initial height of 2.4 meters. How do I write a function and create a graph on a coordinate plane that will model the path that the balloon travelled, the maximum height, the time it took to reach the maximum height, the time when the balloon hit the floor, the time when the balloon is 30 meters, as well as the height of the balloon after 1 second.

Answer 1

Makayla,

I can't really demonstrate the graph in this editor, but the equation is straight forward. The equation for objects in free flight is:

 

y = .5at² + v₀t + y₀   where y is the vertical position of the object, a is the acceleration do to gravity (-9.8 m/s²), t is time elapsed in seconds, v₀ is the initial velocity, and y₀ is the initial height.  So if we fill in what we know:

 

y = .5(-9.8)t² + 27t + 2.4 = -4.9t² + 27t + 2.4

 

For the graph, set the vertical axis to y and the horizontal axis to t. You only need quadrant I because you can't have negative time  or negative height. Just pick values of t to plug in and determine the corresponding y values, then plot.

 

Now, as for the maximum height, we'll examine the equation from an algebra point of view. The equation is clearly quadratic (maximum exponent = 2) and therefore graphs as a parabola. Because the squared term is negative, we know that it opens down. Logically, you picture an object thrown into the air following an arcing path up and then back down, like half a McDonald's sign. The maximum height of the object will occur at the vertex of the parabola. The x coordinate of the vertex can be found with the equation: x = -b/(2a)

 

Or in this case, t = -27/(2*(-4.9)) = 2.755 seconds.

 

So we plug this time into the equation to find the maximum height of the balloon:

 

y_max = -4.9(2.755)² + 27(2.755) + 2.4 = 39.6 meters

 

To find the time when the balloon hits the floor, we have to work backwards. We set y = 0 and solve for t.

 

0 = -4.9t² + 27t + 2.4      use the quadratic formula.

 

t = [-27 ± √(27² - 4(-4.9)(2.4))] / (2*(-4.9))

 

t = [-27 ± √(776.04)] / (-9.8)

 

t = -.087  or   5.60     Again, negative time would not be appropriate, so:

 

t_floor = 5.6 seconds

 

Now for the time when the balloon is at y = 30, we put 30 in for y and solve exactly the same way. However, Since we know the max height is 39.6, the baloon will pass through y = 30 twice. We'll have two valid answers of when.

 

30 = -4.9t² + 27t + 2.4     Set equal to zero

0 = -4.9t² + 27t - 27.6      Use quadratic formula

 

t = [-27 ± √(27² - 4(-4.9)(-27.6))] / (2(-4.9))

 

t = [-27 ± √(188.04)] / (-9.8)

 

t₃₀ = 1.36 seconds and 4.15 seconds

 

The height at one second requires only that you plug in 1 for t

 

y_1sec = -4.9(1)² + 27(1) + 2.4 = 24.5 meters

 

I hope that helps.