a rock is thrown unward off a cliff at a 45 degree angle. THe rock falls back to the elevation it was thrown from after 3.56 seconds. It hits the water 5.31 seconds after it was thrown. How high is the cliff? what is the horizontal distance from the cliff to the point in the water that the rock lands?

Answer 1

consider the vertical movement
hf=hi+Viv*t-1/2 g t^2
first equation: 0=Viv*3.56-4.9(3.56^2)
solve for Viv, the initial componentof vertical velocity

Second equation: hf=hi+Viv*5.31-4.9(5.31)^2
0=hi+ Viv....
solve for hi, the initial height of the cliff.

Finally, initial horizontal velocity (at 45 deg) must be the same magnitude as the initial vertical component.
horizontal distance= Vih*5.31
where Vih=Viv