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ra1l [238]
3 years ago
8

Bonus Points. A ball was thrown straight up in the air from 1.2 meter above the grouind. After 3 seconds the ball returnes to th

e ground. Determine the maximum height it reached.
Physics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

Explanation:

given that

Distance above the ground, s = 1.2 m

Time taken by the ball, t = 3 s

Velocity of the ball, v = 1.2/3 = 0.4 m/s

Maximum height reached by the ball is then given by the formula

H = v² / 2g

H = 0.4² / 2 * 9.8

H = 0.16 / 19.6

H = 0.0082 m or rather, 0.82 cm

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What is the energy source in each picture?
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Answer:

Explanation:

Discussion

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The bottom picture is much harder. T energy is provided by the forefinger. It has to move. If you just stick your finger in the water, all that will happen is that you will move some water (very little in the scheme of things.

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In order to change the color of light you must change
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Answer:

The answer is C either the frequency or the wave lenght of the light

Explanation:

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4 years ago
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3 years ago
What would the dogs velocity be if ran the same distance in the opposite direction but covered the distance in 24 seconds
iVinArrow [24]

Answer:

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7 0
4 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
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