![To determine the velocity of the watermelon right before it hits the ground, we can use the formula for the final velocity of an object in free fall:
\[ v = \sqrt{2gh} \]
where:
- \( v \) is the final velocity,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height from which the object is dropped (8 meters in this case).
Plugging in the values:
\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 8 \, \text{m}} \]
\[ v = \sqrt{157.28}](https://theinternetioimages-f2htc2ggcrath6d7.z01.azurefd.net/$web/askai/b8c81ec9-bebe-4550-bc8f-3119f12d189f.webp)
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Question: A 2.5 Kg watermelon is dropped off an 8-meter balcony. What will the watermelon’s velocity be right before it hits the ground? 196 m/s 12.52 m/s 8.23 m/s 156.8 m/s
To determine the velocity of the watermelon just before it hits the ground, we can use the principles of physics, specifically the equations of motion under constant acceleration (gravity). The equation we will use is: \[ v^2 = u^2 + 2gh \] where: - \( v \) is the final velocity, - \( u \) is the initial velocity (0 m/s, since it's dropped from rest), - \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), - \( h \) is the height (8 meters in this case). Plugging in the values: \[ v^2 = 0 + 2 \cdot 9.8 \cdot 8 \] \[ v^2 = 156.8 \] \[ v = \sqrt{156.8} \] \[ v \approx 12.52 \, \text{m/s} \] So, the correct answer is 12.52 m/s.
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