A stone falls from rest from the top of a cliff,a second stone is thrown do wnward from the same height 2sec later with an initial speed of 30m/s.If both stones hit the ground below simultaneously,how high is the cliff?

Answers

Answer 1

Answer:

The cliff is 12.632 meter high.

Explanation:

Each stone experiments a free fall motion, that is, an uniformly accelerated motion due to gravity. We construct the respective equations of motion for each stone:

First stone

[tex]y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)

Second stone

[tex]y_{2} = y_{o,2} +v_{o,2}\cdot (t-2) +\frac{1}{2}\cdot g\cdot (t-2)^{2}[/tex] (2)

Where:

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Final height of the first and second stone, in meters.

[tex]y_{o,1}[/tex], [tex]y_{o,2}[/tex] - Initial height of the first and second stone, in meters.

[tex]v_{o,1}[/tex], [tex]v_{o,2}[/tex] - Initial speed of the first and second stone, in meters per second.

[tex]t[/tex] - Time, in seconds.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

If we know that [tex]y_{o,1} = y_{o,2}[/tex], [tex]y_{1} = y_{2} = 0\,m[/tex], [tex]v_{o,1} = 0\,\frac{m}{s}[/tex], [tex]v_{o,2} = -30\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then we find that time when both stones hit the ground simultaneously is:

[tex]4.904\cdot t^{2} = -30\cdot (t-2)+4.904\cdot (t-2)^{2}[/tex]

[tex]4.904\cdot t^{2} = -30\cdot t + 60 +4.904\cdot (t^{2}-4\cdot t +4)[/tex]

[tex]-30\cdot t +60 -19.616\cdot t +19.616 = 0[/tex]

[tex]49.616\cdot t = 79.616[/tex]

[tex]t = 1.605\,s[/tex]

The height of the cliff is:

[tex]y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex]

[tex]y_{o,1} = 12.632\,m[/tex]

The cliff is 12.632 meter high.


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Answers

Answer:

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Answers

Answer:

B - a simple circuit with one bulb and two batteries.

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Answers

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Answer:

unbalanced force

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Answer:

Normal force, the puck will move off in a different direction

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Answers

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Satellites are pulled toward Earth by its gravity, even when they are hundreds of kilometres distant. The satellite doesn't fall back to Earth; instead, it travels into orbit above it due to gravity and the velocity it acquired during launch into space.

A satellite that is in a steady circular orbit around the Earth is always falling. It encounters a gravitational pull toward the centre of Earth as it orbits, which forces it to shift course and “fall” towards Earth.

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