I'd be happy to help solve this physics problem! Let's break it down step-by-step:
1. First, let's identify the relevant information:
- Initial velocity (v₀) = 100 m/s (upward)
- Acceleration due to gravity (g) = -9.8 m/s² (downward)
- Initial height (y₀) = 0 m
- Final height (y) = 0 m (when it returns to the ground)
2. We can use the equation for displacement in constant acceleration:
y = y₀ + v₀t + ½at²
3. Substituting our values:
0 = 0 + 100t + ½(-9.8)t²
4. Simplifying:
0 = 100t - 4.9t²
5. Rearranging to standard quadratic form:
4.9t² - 100t = 0
6. Factoring out t:
t(4.9t - 100) = 0
7. Solving for t:
t = 0 or t = 100/4.9 ≈ 20.41 seconds
8. The solution t = 0 represents the initial moment. We're interested in when it returns to the ground, so our answer is t ≈ 20.41 seconds.
Therefore, ignoring air resistance, the stone will take approximately 20.41 seconds to return to the ground.
Note: This total time includes both the upward journey and the downward journey. The stone reaches its maximum height at half this time, about 10.2 seconds.
Rocket
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