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?
options:
$a = 5, b = 0$
$a = 4, b = 0$
$a = 5, b = 3$
$a = 4, b = 3$

Explanation:

Step1: Analyze projectile motion symmetry

Projectile motion is symmetric around the peak. The peak here is at \( t = 3 \) seconds (height = 7 feet). So, the height at \( t = 1 \) should mirror \( t = 5 \), and \( t = 0 \) should mirror \( t = 6 \).

Step2: Find value of \( A \)

At \( t = 5 \), height is 4. Wait, no, wait. Wait, the symmetry: the distance from peak (\( t = 3 \)): \( t = 1 \) is 2 seconds left of peak, \( t = 5 \) is 2 seconds right of peak. Wait, no, \( t = 0 \) to \( t = 3 \): let's check the heights. At \( t = 2 \), height is 6; \( t = 4 \), height is 6 (symmetric around \( t = 3 \)). So \( t = 1 \) should mirror \( t = 5 \). At \( t = 5 \), height is 4? Wait, no, wait the table: \( t = 0 \): 3, \( t = 1 \): \( A \), \( t = 2 \): 6, \( t = 3 \): 7, \( t = 4 \): 6, \( t = 5 \): 4, \( t = 6 \): \( B \). Wait, maybe symmetry around \( t = 3 \): so \( t = 0 \) and \( t = 6 \) should be equal (since 0 and 6 are 3 units from 3), \( t = 1 \) and \( t = 5 \) should be equal, \( t = 2 \) and \( t = 4 \) are equal (which they are, 6). So \( t = 1 \) (A) should equal \( t = 5 \) height? Wait \( t = 5 \) is 4? No, that can't be. Wait, maybe I messed up. Wait, let's check the time intervals. From \( t = 0 \) to \( t = 3 \), the height increases: 3, A, 6, 7. Then decreases: 7, 6, 4, B. Wait, the increase from \( t = 0 \) to \( t = 3 \): let's see the differences. From \( t = 0 \) to \( t = 1 \): 3 to A. \( t = 1 \) to \( t = 2 \): A to 6. \( t = 2 \) to \( t = 3 \): 6 to 7. Then decrease: \( t = 3 \) to \( t = 4 \): 7 to 6 (decrease by 1), \( t = 4 \) to \( t = 5 \): 6 to 4 (decrease by 2)? Wait, no, that's not symmetric. Wait, maybe the symmetry is around the peak, so the height at \( t = 0 \) should equal \( t = 6 \), and \( t = 1 \) equals \( t = 5 \), \( t = 2 \) equals \( t = 4 \). We see \( t = 2 \) (6) equals \( t = 4 \) (6), good. So \( t = 1 \) (A) should equal \( t = 5 \) (4)? No, that would make A = 4, but then \( t = 0 \) (3) should equal \( t = 6 \) (B), so B = 3. Wait, but let's check the options. One of the options is \( A = 5, B = 3 \)? No, wait the option \( A = 5, B = 3 \)? Wait no, the option \( A = 5, B = 3 \) or \( A = 4, B = 3 \)? Wait, let's re - evaluate. Wait, from \( t = 0 \) (3) to \( t = 1 \) (A) to \( t = 2 \) (6) to \( t = 3 \) (7). The increase: 3 to A: let's say the difference from 0 to 1 is \( A - 3 \), 1 to 2 is \( 6 - A \), 2 to 3 is \( 7 - 6 = 1 \). Then from 3 to 4: \( 6 - 7=- 1 \), 4 to 5: \( 4 - 6=-2 \), 5 to 6: \( B - 4 \). But due to symmetry, the path up and down should mirror. So the increase from 0 to 3 should mirror the decrease from 3 to 6. So the height at \( t = 0 \) (3) should equal \( t = 6 \) (B), so \( B = 3 \). Then, the height at \( t = 1 \) (A) should equal the height at \( t = 5 \) (4)? Wait no, \( t = 5 \) is 4, \( t = 1 \) would be 4? But then \( t = 0 \) is 3, \( t = 1 \) is 4, \( t = 2 \) is 6, \( t = 3 \) is 7, \( t = 4 \) is 6, \( t = 5 \) is 4, \( t = 6 \) is 3. But let's check the options. The option \( A = 5, B = 3 \): no, \( t = 5 \) is 4, so A can't be 5. Wait, wait maybe I made a mistake. Wait the table: \( t = 0 \): 3, \( t = 1 \): A, \( t = 2 \): 6, \( t = 3 \): 7, \( t = 4 \): 6, \( t = 5 \): 4, \( t = 6 \): B. Let's calculate the differences between consecutive times. From \( t = 0 \) to \( t = 1 \): \( A - 3 \). \( t = 1 \) to \( t = 2 \): \( 6 - A \). \( t = 2 \) to \( t = 3 \): \( 7 - 6 = 1 \). \( t = 3 \) to \( t = 4 \): \( 6 - 7=-1 \). \( t = 4 \) to \( t = 5 \): \( 4 - 6=-2 \). \( t = 5 \) to \( t = 6 \): \( B - 4 \). For projectile motion, the acceleration is constant (gravity), so the vertical velocity changes linearly. But maybe the symmetry is that the height at \( t = 0 \) is equal to \( t = 6 \) (since the total time is 6 seconds, so the start and end heights should be equal if it's a throw and catch at the same height). Wait, Mary throws it, her friend catches it. So the initial height (t=0) and final height (t=6) should be the same. So \( B = 3 \) (since t=0 is 3). Then, the height at t=1 (A) should be equal to the height at t=5? Wait t=5 is 4, but that would mean A=4. But wait, let's check the middle. Wait, from t=0 to t=3: 3, A, 6, 7. From t=3 to t=6: 7, 6, 4, 3. Wait, the decrease from t=3 to t=4 is 1 (7 to 6), t=4 to t=5 is 2 (6 to 4), t=5 to t=6 is 1 (4 to 3). The increase from t=0 to t=1 should be 1 (3 to 4), t=1 to t=2 is 2 (4 to 6), t=2 to t=3 is 1 (6 to 7). That matches the decrease (1, 2, 1). So A=4, B=3? Wait no, the option with A=4, B=3 is there? Wait the options: the last option is \( A = 4, B = 3 \). Wait, but let's check again. Wait, t=0:3, t=1:A, t=2:6, t=3:7, t=4:6, t=5:4, t=6:B. If B=3 (same as t=0), then A should be equal to t=5? No, t=5 is 4, so A=4. Then the heights are 3,4,6,7,6,4,3. Which is symmetric around t=3. Yes, that makes sense. Wait, but earlier I thought t=1 and t=5 should be equal, which they are (4), and t=0 and t=6 are equal (3), t=2 and t=4 are equal (6). So A=4, B=3.

Answer:

\( A = 4, B = 3 \) (the option: \( A = 4, B = 3 \))