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 is symmetric about the vertex (highest point). The highest point occurs at \(t = 3\) seconds (\(y=7\) feet).

Step2: Find the value of \(A\)

For \(t = 1\) and \(t = 5\), the heights are symmetric. When \(t = 5\), \(y = 4\). So when \(t=1\), \(A = 4\) (because of symmetry \(y(1)=y(5)\))

Step3: Find the value of \(B\)

For \(t = 0\) and \(t = 6\), the heights are symmetric. When \(t = 0\), \(y = 3\). But wait, no! Wait, actually, the initial height \(y(0)=3\). But for projectile motion, if we consider the general form \(y = ax^{2}+bx + c\) (where \(x=t\)). Substituting \(t = 0,y=3\) gives \(c = 3\). Substituting \(t=2,y = 6\): \(6=4a + 2b+3\) (i.e., \(4a+2b=3\)). Substituting \(t = 3,y = 7\): \(7 = 9a+3b + 3\) (i.e., \(9a+3b=4\)). Solving the system \(

$$\begin{cases}4a + 2b=3\\9a+3b=4\end{cases}$$

\). Multiply the first equation by \(3\): \(12a+6b = 9\). Multiply the second equation by \(2\): \(18a+6b=8\). Subtract: \(- 6a=1\), \(a=-\frac{1}{6}\). Then from \(4a+2b=3\), \(4\times(-\frac{1}{6})+2b=3\), \(-\frac{2}{3}+2b=3\), \(2b=\frac{11}{3}\), \(b=\frac{11}{6}\). The equation is \(y=-\frac{1}{6}t^{2}+\frac{11}{6}t + 3\). When \(t = 6\), \(y=-\frac{1}{6}\times36+\frac{11}{6}\times6+3=-6 + 11+3=8\)? No, wait, another approach. Since it's caught at \(t = 6\). The height at \(t = 0\) and \(t = 6\) (assuming it's thrown from height \(y(0)\) and caught at \(y(6)\)). But wait, no, actually, the symmetry about \(t = 3\). The difference between \(t = 0\) and \(t=3\) is \(3\) units, and between \(t=3\) and \(t = 6\) is \(3\) units. The height at \(t=0\) is \(3\). But wait, no, using the table values. The height at \(t = 2\) is \(6\), at \(t=4\) is \(6\) (symmetry about \(t = 3\)). The height at \(t=1\) and \(t = 5\): at \(t=5,y = 4\), so at \(t=1,y=4\). At \(t=0,y = 3\), and since the motion is projectile (a parabola \(y=ax^{2}+bx + c\)), when \(t = 6\), \(y=3\) (because of the symmetry of the parabola \(y(t)=y(6 - t)\) for the given data points in the table. Wait, no, looking at the values: \(t = 0,y=3\); \(t=1,y = A\); \(t=2,y=6\); \(t=3,y = 7\); \(t=4,y=6\); \(t=5,y = 4\); \(t=6,y=B\). The pattern of heights is symmetric about \(t = 3\). So \(A\) (at \(t = 1\)) has the same value as at \(t=5\) (which is \(4\)), and \(B\) (at \(t = 6\)) has the same value as at \(t=0\) (which is \(3\))? No, wait, no! Wait, the function \(y(t)\) for projectile motion (a quadratic function \(y=at^{2}+bt + c\)) passing through \((0,3),(2,6),(3,7)\). We found \(y=-\frac{1}{6}t^{2}+\frac{11}{6}t + 3\). When \(t = 6\), \(y=-\frac{1}{6}\times36+\frac{11}{6}\times6+3=- 6+11 + 3=8\)? No, wrong. Wait, looking at the symmetry of the table values (not the physics formula). The height at \(t=1\) and \(t = 5\): since \(t=2\) and \(t=4\) have the same height (\(6\)), \(t=1\) and \(t=5\) should have the same height. Given \(t = 5,y = 4\), so \(A = 4\). The height at \(t=0\) and \(t=6\) (if we assume the motion is symmetric about \(t = 3\)). But \(t=0\) is \(3\) units before \(t = 3\) and \(t=6\) is \(3\) units after \(t = 3\). So \(B=3\)

Answer:

\(A = 4,B = 3\) (i.e., the option \(A = 4,B = 3\))