QuestionMay 6, 2025

17. Clay drops a water balloon from a 20-foot platform onto Bailey's head who is four feet tall. How long does it take for the water balloon to reach Bailey's head?

17. Clay drops a water balloon from a 20-foot platform onto Bailey's head who is four feet tall. How long does it take for the water balloon to reach Bailey's head?
17. Clay drops a water balloon from a 20-foot platform onto Bailey's head who is four feet tall. How long does it take for the water balloon to reach Bailey's head?

Solution
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Answer

1 \, \text{second} Explanation 1. Write the equation of motion Use the formula for free fall: h = \frac{1}{2} g t^2, where h is the height fallen, g = 32 \, \text{ft/s}^2 (acceleration due to gravity), and t is time. The balloon falls from a height of 20 - 4 = 16 \, \text{ft}. 2. Solve for time t Rearrange the formula: t = \sqrt{\frac{2h}{g}}. Substituting h = 16 and g = 32: t = \sqrt{\frac{2(16)}{32}} = \sqrt{1} = 1 \, \text{s}.

Explanation

1. Write the equation of motion<br /> Use the formula for free fall: $h = \frac{1}{2} g t^2$, where $h$ is the height fallen, $g = 32 \, \text{ft/s}^2$ (acceleration due to gravity), and $t$ is time. The balloon falls from a height of $20 - 4 = 16 \, \text{ft}$.<br /><br />2. Solve for time $t$<br /> Rearrange the formula: $t = \sqrt{\frac{2h}{g}}$. Substituting $h = 16$ and $g = 32$: <br />$t = \sqrt{\frac{2(16)}{32}} = \sqrt{1} = 1 \, \text{s}$.
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