Eksperimentti: hyppykorkeuden määrittäminen lentoajalla. - Revision history

Mol: /* Example */

2022-05-03T16:12:03Z

Example

← Older revision		Revision as of 19:12, 3 May 2022
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	In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.		In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.

	The takeoff velocity is		The takeoff velocity is <math>v_0 = gt_m = 9.81 m/s^2 \times 0.16 s = 1.5696 m/s</math>.

2022-05-03T16:11:06Z

Example

← Older revision		Revision as of 19:11, 3 May 2022
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	In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.		In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.

	~~Also take-off~~ velocity is ~~easy to calculate as a difference.~~		The takeoff velocity is

2022-05-03T16:10:30Z

Theory

← Older revision		Revision as of 19:10, 3 May 2022
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	The time at the maximum height (<math>t_m</math>) will be half of the flight time.		The time at the maximum height (<math>t_m</math>) will be half of the flight time.

			== Example ==

	In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.		In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.

	Also take-off velocity is easy to calculate as a difference.		Also take-off velocity is easy to calculate as a difference.

2022-05-03T15:05:52Z

Theory

← Older revision		Revision as of 18:05, 3 May 2022
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	[[File:ForcePlate jumping2 timeOfFlight.svg\|thumb]]		[[File:ForcePlate jumping2 timeOfFlight.svg\|thumb]]

	<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt_m</math> ~~which~~. Combining these two we have		<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt_m</math>. Combining these two we have

	<math>		<math>

2022-05-03T14:53:24Z

Theory

← Older revision		Revision as of 17:53, 3 May 2022
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	== Theory ==		== Theory ==

			[[File:ForcePlate jumping2 timeOfFlight.svg\|thumb]]

	<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt_m</math> which. Combining these two we have		<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt_m</math> which. Combining these two we have
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	The time at the maximum height (<math>t_m</math>) will be half of the flight time.		The time at the maximum height (<math>t_m</math>) will be half of the flight time.

			In the experiment, we have <math>t_m = 0.16</math> s, which gives <math> h = \tfrac12 gt_m^2 = \frac12 \times 9.81 \times 0.16^2 = 0.13</math> m.

	Also take-off velocity is easy to calculate as a difference.		Also take-off velocity is easy to calculate as a difference.

2022-05-03T14:45:18Z

Theory

← Older revision		Revision as of 17:45, 3 May 2022
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	\end{matrix}		\end{matrix}
	</math>		</math>

	The time at the maximum height (<math>t_m</math>) will be half of the flight time.		The time at the maximum height (<math>t_m</math>) will be half of the flight time.


	Also take-off velocity is easy to calculate as a difference.		Also take-off velocity is easy to calculate as a difference.

2022-05-03T14:45:02Z

Theory

← Older revision		Revision as of 17:45, 3 May 2022
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	h &= v_0 t - \tfrac12 gt^2 \\		h &= v_0 t - \tfrac12 gt^2 \\
	&= gt_m^2 - \tfrac12 gt_m^2 \\		&= gt_m^2 - \tfrac12 gt_m^2 \\
			&= \tfrac12 gt_m^2
	\end{matrix}		\end{matrix}
	</math>		</math>
			The time at the maximum height (<math>t_m</math>) will be half of the flight time.

	~~<math>h = \frac{v^2}{2g}</math>~~

	Also take-off velocity is easy to calculate as a difference.		Also take-off velocity is easy to calculate as a difference.

2022-05-03T14:43:18Z

Theory

← Older revision		Revision as of 17:43, 3 May 2022
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	<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt</math> which. Combining these two we have		<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt_m</math> which. Combining these two we have

	<math>		<math>
	\begin{matrix}		\begin{matrix}
	h &= v_0 t - \tfrac12 gt^2 \\		h &= v_0 t - \tfrac12 gt^2 \\
	&= gt^2 - \tfrac12 gt^2 \\		&= gt_m^2 - \tfrac12 gt_m^2 \\
	\end{matrix}		\end{matrix}
	</math>		</math>

2022-05-03T14:42:32Z

Theory

2022-05-03T14:40:01Z

Theory

← Older revision		Revision as of 17:40, 3 May 2022
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	<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>.		<math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt</math> which. Combining these two we have

	However, at the maximum height we have that <math>v=0</math>, and thus <math>v = ~~v_0 -~~ gt</math>

			<math>
			h &= v_0 t - \tfrac12 gt^2 \\
			&= gt^2 - \tfrac12 gt^2 \\
			</math>

	<math>h = \frac{v^2}{2g}</math>		<math>h = \frac{v^2}{2g}</math>

	Also take-off velocity is easy to calculate as a difference.		Also take-off velocity is easy to calculate as a difference.