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? time (seconds) height (feet) 0 3 1 a 2 6 3 7 4 6 5 4 6 b a = 5, b = 3 a = 4, b = 0 a = 4, b = 3 a = 5, b = 0

Explanation:

Step1: Analyze projectile - motion symmetry

Projectile motion is symmetric about the vertex. The time from the start to the vertex and from the vertex to the end (when it returns to the same height) is the same. The disc reaches its maximum height at \(t = 3\) seconds.

Step2: Determine value of \(A\)

Since the motion is symmetric, the height at \(t = 1\) second is the same as the height at \(t = 5\) seconds. Given the height at \(t = 5\) seconds is 4 feet, so \(A=4\).

Step3: Determine value of \(B\)

The disc is caught at \(t = 6\) seconds. At the start (\(t = 0\)) the height is 3 feet. Since the motion starts and ends at the same - height level (assuming the friend catches it at the same height Mary throws it from in the vertical - direction), \(B = 0\).

Answer:

B. \(A = 4, B = 0\)