Question

1. serge and a friend are throwing a paper airplane to each other. they stand 5 m apart from each other and catch the airplane at a height of 1 m above the ground. serge throws the airplane on a parabolic flight path that achieves a minimum height of 0.5 m halfway to his friend.

a) determine a quadratic function that models the flight path for the height of the airplane.
b) determine the height of the plane when it is a horizontal distance of 1 m from serge’s friend.
c) state the domain and range of the function.

Explanation:

Step1: Identify the vertex and points

The parabola has a minimum (since it's a parabolic flight path with a minimum height) at the midpoint of the horizontal distance between Serge and his friend. The horizontal distance between them is 5 m, so the midpoint (vertex's x - coordinate) is at \(x=\frac{5}{2}=2.5\) m. The minimum height (vertex's y - coordinate) is \(y = 0.5\) m. The starting point (Serge's position) is at \(x = 0\), \(y=1\) m (since he catches it at 1 m above the ground) and the ending point (friend's position) is at \(x = 5\), \(y = 1\) m.

The vertex form of a quadratic function is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex. Here, \(h = 2.5\), \(k=0.5\), so the equation is \(y=a(x - 2.5)^{2}+0.5\).

We can use the point \((0,1)\) to find \(a\). Substitute \(x = 0\) and \(y = 1\) into the equation:

\(1=a(0 - 2.5)^{2}+0.5\)

\(1=a(6.25)+0.5\)

Subtract 0.5 from both sides: \(0.5 = 6.25a\)

Then \(a=\frac{0.5}{6.25}=\frac{1}{12.5}=0.08\)

So the quadratic function is \(y = 0.08(x - 2.5)^{2}+0.5\) or expanding it:

\(y=0.08(x^{2}-5x + 6.25)+0.5=0.08x^{2}-0.4x+0.5 + 0.5=0.08x^{2}-0.4x + 1\)

Step2: Calculate height at 1 m from Serge

Serge is at \(x = 0\). 1 m from Serge means \(x=1\). Substitute \(x = 1\) into the quadratic function \(y = 0.08(1 - 2.5)^{2}+0.5\)

First, calculate \((1 - 2.5)=- 1.5\), then \((-1.5)^{2}=2.25\)

Then \(0.08\times2.25 = 0.18\)

\(y=0.18 + 0.5=0.68\) m? Wait, no, wait. Wait, the starting point is at \(x = 0\), \(y = 1\). Wait, maybe I made a mistake in the vertex. Wait, the problem says "achieves a minimum height of 0.5 m halfway to his friend". The distance between them is 5 m, so halfway is 2.5 m from Serge. So the vertex is at \(x = 2.5\), \(y = 0.5\). The two endpoints are \((0,1)\) and \((5,1)\). Let's re - calculate \(a\) correctly.

Using \((0,1)\) in \(y=a(x - 2.5)^{2}+0.5\):

\(1=a(0 - 2.5)^{2}+0.5\)

\(1=a(6.25)+0.5\)

\(a=\frac{1 - 0.5}{6.25}=\frac{0.5}{6.25}=\frac{1}{12.5}=0.08\). So the function is correct. Now, when \(x = 1\) (1 m from Serge, who is at \(x = 0\)):

\(y=0.08(1 - 2.5)^{2}+0.5=0.08\times(-1.5)^{2}+0.5=0.08\times2.25 + 0.5=0.18+0.5 = 0.68\)? But that can't be, because at \(x = 0\), \(y = 1\), and as we move towards the midpoint (x = 2.5), the height should decrease. Wait, 1 m from Serge is \(x = 1\), which is between \(x = 0\) and \(x = 2.5\), so the height should be between 1 m and 0.5 m. Wait, my calculation gives 0.68, which is between 0.5 and 1. Let's check with the expanded form: \(y=0.08x^{2}-0.4x + 1\). When \(x = 1\), \(y=0.08(1)-0.4(1)+1=0.08 - 0.4 + 1=0.68\) m.

Step3: Determine domain and range

The domain is the set of all possible x - values (horizontal distances). The airplane moves from \(x = 0\) (Serge's position) to \(x = 5\) (friend's position), so domain is \(0\leq x\leq5\) or \([0,5]\) in interval notation.

The range is the set of all possible y - values (heights). The minimum height is 0.5 m and the maximum height is 1 m (at the endpoints \(x = 0\) and \(x = 5\)), so range is \(0.5\leq y\leq1\) or \([0.5,1]\) in interval notation.

Answer:

a) The quadratic function is \(y = 0.08(x - 2.5)^{2}+0.5\) (or \(y=0.08x^{2}-0.4x + 1\))

b) The height when \(x = 1\) is \(0.68\) m (or using the vertex form: \(y = 0.08(1 - 2.5)^{2}+0.5=0.68\) m)

c) Domain: \([0,5]\), Range: \([0.5,1]\)