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:
ymax = -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:
tfloor = 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
y1sec = -4.9(1)2 + 27(1) + 2.4 = 24.5 meters
I hope that helps.