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A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the time in seconds at which the stone reaches its maximum height.
g =9.8 m / sec2
- a)1.67
- b)2.7
- c)2.04
- d)2.8
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A stone thrown from the top of a building is given an initial velocity...
Explanation:
Initial velocity u = 20.0 m/s
At maximum height it ll stop
So final velocity v = 0 m/s
Acceleration due to gravity g = 9.8 m/s2
Time taken to reach maximum height = t
We know
v = u + at
Most Upvoted Answer
A stone thrown from the top of a building is given an initial velocity...
To determine the time at which the stone reaches its maximum height, we can use the kinematic equation for vertical motion:
\(v_f = v_i + at\),
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
In this case, the stone is thrown straight upward, so the acceleration is equal to the acceleration due to gravity, \(a = -9.8 \, \text{m/s}^2\) (negative because it acts in the opposite direction to the initial velocity).
The initial velocity is given as 20.0 m/s, and the final velocity at the maximum height is 0 m/s (because the stone momentarily stops before falling back down). Therefore, we can rewrite the equation as:
\(0 \, \text{m/s} = 20.0 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot t\).
Simplifying this equation, we have:
\(9.8 \, \text{m/s}^2 \cdot t = 20.0 \, \text{m/s}\).
To find \(t\), we divide both sides of the equation by \(9.8 \, \text{m/s}^2\):
\(t = \dfrac{20.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 2.04 \, \text{s}\).
Therefore, the time at which the stone reaches its maximum height is approximately 2.04 seconds.
Therefore, the correct answer is option 'C' (2.04 seconds).
\(v_f = v_i + at\),
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
In this case, the stone is thrown straight upward, so the acceleration is equal to the acceleration due to gravity, \(a = -9.8 \, \text{m/s}^2\) (negative because it acts in the opposite direction to the initial velocity).
The initial velocity is given as 20.0 m/s, and the final velocity at the maximum height is 0 m/s (because the stone momentarily stops before falling back down). Therefore, we can rewrite the equation as:
\(0 \, \text{m/s} = 20.0 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot t\).
Simplifying this equation, we have:
\(9.8 \, \text{m/s}^2 \cdot t = 20.0 \, \text{m/s}\).
To find \(t\), we divide both sides of the equation by \(9.8 \, \text{m/s}^2\):
\(t = \dfrac{20.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 2.04 \, \text{s}\).
Therefore, the time at which the stone reaches its maximum height is approximately 2.04 seconds.
Therefore, the correct answer is option 'C' (2.04 seconds).
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A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the time in seconds at which the stone reaches its maximum height.g =9.8 m / sec2a)1.67b)2.7c)2.04d)2.8Correct answer is option 'C'. Can you explain this answer?
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