Question

step 8: observe how changes in the speed of the bottle affect beanbag height
calculate the average maximum height for all three trials when the speed of the bottle is 2 m/s, 3 m/s, 4 m/s, 5 m/s, and 6 m/s.
record your calculations in table b of your student guide.
when the speed of the bottle is 2 m/s, the average maximum height of the beanbag is 0.31 m.
when the speed of the bottle is 3 m/s, the average maximum height of the beanbag is dropdown m.
when the speed of the bottle is 4 m/s, the average maximum height of the
when the speed of the
maximum height of the
when the speed of the
maximum height of the
dropdown options: 0.15, 0.42, 0.43, 0.44

Explanation:

To determine the average maximum height of the beanbag when the bottle speed is 3 m/s, we analyze the relationship between speed and height (kinetic energy to potential energy conversion, \( KE = PE \) or \( \frac{1}{2}mv^2 = mgh \), so \( h \propto v^2 \)).

Step 1: Analyze the proportionality

From the formula \( h = \frac{v^2}{2g} \) (derived from \( \frac{1}{2}mv^2 = mgh \), cancel \( m \), solve for \( h \)), height \( h \) is proportional to \( v^2 \).

Step 2: Calculate the ratio of speeds squared

For \( v_1 = 2 \, \text{m/s} \) (height \( h_1 = 0.31 \, \text{m} \)) and \( v_2 = 3 \, \text{m/s} \):
\[
\frac{v_2^2}{v_1^2} = \frac{3^2}{2^2} = \frac{9}{4} = 2.25
\]

Step 3: Find the expected height for \( v = 3 \, \text{m/s} \)

Multiply \( h_1 \) by the ratio:
\[
h_2 = h_1 \times 2.25 = 0.31 \times 2.25 \approx 0.6975
\] Wait, this conflicts with the given options. Re - evaluate: maybe the data follows a pattern where height increases with speed, and from the options (0.15, 0.42, 0.43, 0.44), 0.42 is the most reasonable next value after 0.31 as speed increases (2 m/s → 3 m/s, height should increase, and 0.42 is a logical increment).

Answer:

0.42