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

12.25m/s

Step-by-step explanation:

`d=v_ot+(1)/(2)at^2`

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

`d=(1)/(2)(9.8)(3)^2=44.1m`

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

`44.1=v_o(2)+(1)/(2)(9.8)(2)^2 \\\\\\44.1=2v_o+19.6 \\\\\\2v_o=24.5 \\\\\\v_o=12.25m/s`

Hope this helps!