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ectile \( \begin{array}{l}\text { 19. An object moving horizontally with speed } v \text { falls off } \\ \text { the edge of a vertical cliff and lands a distance } d \\ \text { from the base of the cliff. How far from the base of } \\ \text { the cliff would the object land if the height of the } \\ \text { cliff was doubled? } \\ \begin{array}{ll}\text { A) } d & \text { B) } \sqrt{2} d \\ \text { C) } 2 d & \text { D) } 2 \sqrt{2} d \\ \text { E) } 4 d & \end{array}\end{array}> \).

Ask by Ruiz Lee. in the United States
Feb 04,2025

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Answer

If the height of the cliff is doubled, the object will land \( \sqrt{2} \) times farther from the base. So, the answer is \( \sqrt{2}\, d \).

Solution

To determine how far the object would land if the height of the cliff is doubled, let's analyze the problem step by step. ### Original Scenario: - **Initial Height**: \( h \) - **Horizontal Speed**: \( v \) - **Time to Fall (\( t \))**: \[ t = \sqrt{\frac{2h}{g}} \] - **Horizontal Distance (\( d \))**: \[ d = v \cdot t = v \cdot \sqrt{\frac{2h}{g}} \] ### New Scenario (Height Doubled): - **New Height**: \( 2h \) - **New Time to Fall (\( t' \))**: \[ t' = \sqrt{\frac{2 \cdot 2h}{g}} = \sqrt{\frac{4h}{g}} = 2 \cdot \sqrt{\frac{h}{g}} = \sqrt{2} \cdot \sqrt{\frac{2h}{g}} = \sqrt{2} \cdot t \] - **New Horizontal Distance (\( d' \))**: \[ d' = v \cdot t' = v \cdot \sqrt{2} \cdot t = \sqrt{2} \cdot (v \cdot t) = \sqrt{2} \cdot d \] ### Conclusion: If the height of the cliff is doubled, the horizontal distance where the object lands becomes \( \sqrt{2} \) times the original distance \( d \). **Answer:** **B) \( \sqrt{2}\, d \)**

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When an object falls off a cliff, its horizontal motion is independent of its vertical motion. The time it takes to hit the ground depends solely on the height of the cliff and is given by the equation \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height and \( g \) is the acceleration due to gravity. If the height of the cliff is doubled, the time to fall becomes \( t = \sqrt{\frac{2(2h)}{g}} = \sqrt{2} \times \sqrt{\frac{2h}{g}} \). Thus, the object will spend more time in the air and travel further horizontally. If the horizontal speed \( v \) remains the same, and the fall takes \( \sqrt{2} \times t_1 \) (where \( t_1 \) is the original time), the new distance \( d' = v \times (\sqrt{2} \times t_1) = \sqrt{2} d \). Hence, the new distance from the base of the cliff is \( \sqrt{2} d \). So the answer is \( \text{B) } \sqrt{2} d \).

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