Question

you will drop the bottle/water mass so that it hits the lever at different speeds. since an object in free fall is accelerated by gravity, you need to determine the heights necessary to drop the bottle to achieve the speeds of 2 m/s, 3 m/s, 4 m/s, 5 m/s, and 6 m/s. use the equation $ht = \frac{v^2}{2g}$ to calculate the height, where $ht$ is the height, $v$ is the speed (velocity), and $g$ is the gravitational acceleration of 9.8 m/s². record these heights in table b. to achieve a speed of 2 m/s, the bottle must be dropped at \\( \boldsymbol{\\_\\_\\_} \\) m. to achieve a speed of 3 m/s, the bottle must be dropped at \\( \boldsymbol{\\_\\_\\_} \\) m. to achieve a speed of 4 m/s, the bottle must be dropped at \\( \boldsymbol{\\_\\_\\_} \\) m. to achieve a speed of 5 m/s, the bottle must be dropped at \\( \boldsymbol{\\_\\_\\_} \\) m. to achieve a speed of 6 m/s, the bottle must be dropped at \\( \boldsymbol{\\_\\_\\_} \\) m.

Explanation:

Step1: For speed \( v = 2 \, \text{m/s} \)

Use the formula \( Ht=\frac{v^{2}}{2g} \), where \( v = 2 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( Ht=\frac{2^{2}}{2\times9.8}=\frac{4}{19.6}\approx0.204 \, \text{m} \)

Step2: For speed \( v = 3 \, \text{m/s} \)

Use the formula \( Ht=\frac{v^{2}}{2g} \), where \( v = 3 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( Ht=\frac{3^{2}}{2\times9.8}=\frac{9}{19.6}\approx0.459 \, \text{m} \)

Step3: For speed \( v = 4 \, \text{m/s} \)

Use the formula \( Ht=\frac{v^{2}}{2g} \), where \( v = 4 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( Ht=\frac{4^{2}}{2\times9.8}=\frac{16}{19.6}\approx0.816 \, \text{m} \)

Step4: For speed \( v = 5 \, \text{m/s} \)

Use the formula \( Ht=\frac{v^{2}}{2g} \), where \( v = 5 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( Ht=\frac{5^{2}}{2\times9.8}=\frac{25}{19.6}\approx1.276 \, \text{m} \)

Step5: For speed \( v = 6 \, \text{m/s} \)

Use the formula \( Ht=\frac{v^{2}}{2g} \), where \( v = 6 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( Ht=\frac{6^{2}}{2\times9.8}=\frac{36}{19.6}\approx1.837 \, \text{m} \)

Answer:

• For \( v = 2 \, \text{m/s} \): \( \approx 0.20 \, \text{m} \) (or \( 0.204 \, \text{m} \))
• For \( v = 3 \, \text{m/s} \): \( \approx 0.46 \, \text{m} \) (or \( 0.459 \, \text{m} \))
• For \( v = 4 \, \text{m/s} \): \( \approx 0.82 \, \text{m} \) (or \( 0.816 \, \text{m} \))
• For \( v = 5 \, \text{m/s} \): \( \approx 1.28 \, \text{m} \) (or \( 1.276 \, \text{m} \))
• For \( v = 6 \, \text{m/s} \): \( \approx 1.84 \, \text{m} \) (or \( 1.837 \, \text{m} \))