Question

mary throws a plastic disc to her friend. her friend catches the disc six seconds after mary throws it. the table shows the height of the disc at one - second intervals. assuming that the throw represents projectile motion, what are the missing values in the table? a = 5, b = 3 a = 4, b = 0 a = 4, b = 3 a = 5, b = 0

Explanation:

Step1: Analyze the symmetry of projectile motion

Projectile motion has symmetry about the vertex (highest - point). The highest - point occurs at \(t = 3\) seconds (height \(y=7\) feet).

Step2: Find the value of \(A\)

For \(t = 1\) and \(t = 5\), using the symmetry of the parabola (since projectile motion follows a parabolic path \(y = ax^{2}+bx + c\)). The height at \(t = 1\) and \(t = 5\) should satisfy the symmetry. We know that if \(y(t)=ax^{2}+bx + c\), then \(y(1)=y(5)\). Given \(y(5) = 4\), so \(A = 4\).

Step3: Find the value of \(B\)

At \(t = 0\), \(y(0)=3\) (initial height). Due to the symmetry of the projectile motion (the motion starts at \(y(0)\) and ends at \(y(6)\)). Since the motion is symmetric about \(t = 3\), and the initial height \(y(0)=3\), then \(B=y(6)\). Also, we can think of the general form of a parabola \(y=a(x - h)^{2}+k\) (vertex form, where \((h,k)=(3,7)\)). Substituting \(x = 0,y = 3\) into \(y=a(x - 3)^{2}+7\), we get \(3=a(0 - 3)^{2}+7\), \(9a=- 4\), \(a=-\frac{4}{9}\). Then \(y=-\frac{4}{9}(x - 3)^{2}+7\). When \(x = 6\), \(y=-\frac{4}{9}(6 - 3)^{2}+7=-\frac{4}{9}\times9 + 7=-4 + 7=3\). But another way: since the motion is symmetric about \(t = 3\), and the disc is thrown from \(y(0) = 3\), it will return to the same height (assuming no air - resistance, which is the case for ideal projectile motion) at \(t = 6\). So \(B = 3\).

Answer:

\(A = 4,B = 3\) (Option \(A = 4,B = 3\))