Question

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?

Answer 1

Answer:

12.25m/s

Explanation:

$d=v_ot+\dfrac{1}{2}at^2$

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

$d=\dfrac{1}{2}(9.8)(3)^2=44.1m$

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

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

Hope this helps!