Question
If a coin is thrown upwards from a height of 3.1 m, with an initial speed of 1 m/s, how many seconds does it remain in the air?
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Answer to a math question If a coin is thrown upwards from a height of 3.1 m, with an initial speed of 1 m/s, how many seconds does it remain in the air?
Velda
4.5110 Answers
To determine how long the coin remains in the air, we need to find the time it takes to reach a height of 0 m (when it falls back down).
The height of the coin at any given time (t) can be determined by the equation:
h(t) = h_0 + v_0t - \frac{1}{2}gt^2
where:
- h(t)
- h_0 = 3.1 \, \text{m}
- v_0 = 1 \, \text{m/s}
- g = 9.81 \, \text{m/s}^2
At the highest point, the vertical speed becomes 0, so:
v(t) = v_0 - gt = 0
Solving for t:
1 - 9.81t = 0
t = \frac{1}{9.81} \approx 0.102 \, \text{s}
The total time in the air is twice this time since it takes the same time to go up and to come down:
T_{\text{total}} = 2 \times 0.102 \, \text{s} = 0.204 \, \text{s}
\boxed{0.204 \, \text{s}}
The height of the coin at any given time (t) can be determined by the equation:
where:
-
-
-
-
At the highest point, the vertical speed becomes 0, so:
Solving for t:
The total time in the air is twice this time since it takes the same time to go up and to come down:
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