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.

time (seconds)	height (feet)
1	a
2	6
3	7
4	6
5	4
6	b

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 projectile motion symmetry

Projectile motion is symmetric, so the height at time \( t \) and \( 6 - t \) should be equal (since total time is 6s). At \( t = 0 \), height is 3. So at \( t = 6 \), height \( B \) should equal height at \( t = 0 \), so \( B = 3 \)? Wait, no, wait. Wait, let's check the symmetry around the peak. The peak is at \( t = 3 \) (since height at \( t = 3 \) is 7, which is the maximum). So the height at \( t = 1 \) should equal height at \( t = 5 \)? Wait, no, \( t = 1 \) and \( t = 5 \): height at \( t = 5 \) is 4. Wait, maybe I made a mistake. Wait, let's list the times: 0,1,2,3,4,5,6. The peak is at \( t = 3 \). So the height at \( t = 0 \) should equal height at \( t = 6 \) (since 0 and 6 are equidistant from 3: \( 3 - 0 = 3 \), \( 6 - 3 = 3 \)). Height at \( t = 0 \) is 3, so \( B = 3 \). Now, height at \( t = 1 \) should equal height at \( t = 5 \)? Wait, \( t = 1 \) and \( t = 5 \): \( 3 - 1 = 2 \), \( 5 - 3 = 2 \). Height at \( t = 5 \) is 4, so \( A = 4 \)? Wait, no, wait \( t = 2 \) and \( t = 4 \): height at \( t = 2 \) is 6, height at \( t = 4 \) is 6, which matches symmetry (distance from 3 is 1 for both). So \( t = 1 \) (distance 2 from 3) and \( t = 5 \) (distance 2 from 3): height at \( t = 5 \) is 4, so \( A = 4 \). And \( t = 0 \) and \( t = 6 \) (distance 3 from 3): height at \( t = 0 \) is 3, so \( B = 3 \). So \( A = 4 \), \( B = 3 \).

Step2: Verify with options

Looking at the options, \( A = 4 \), \( B = 3 \) is one of the options (third option: \( A = 4, B = 3 \)).

Answer:

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