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A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff, A second rock is thrown vertically from the cliff, and it takes this rock 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock thrown, assuming no air resistance?

Respuesta :

osikbro

Answer:

12.25m/s

Explanation:

[tex]d=v_ot+\dfrac{1}{2}at^2[/tex]

Since the initial velocity of the dropped rock is 0, you can write this as:

[tex]d=\dfrac{1}{2}(9.8)(3)^2=44.1m[/tex]

Now, you can set up the equation for the thrown rock:

[tex]44.1=v_o(2)+\dfrac{1}{2}(9.8)(2)^2 \\\\\\44.1=2v_o+19.6 \\\\\\2v_o=24.5 \\\\\\v_o=12.25m/s[/tex]

Hope this helps!

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