a ball rolls of from a cliff with a downward angle of 35 degrees. After 0.710seconds it lands 13m from the cliff. What is its velocity before it hits the ground?

Answer 1

The horizontal velocity is 13m/0.71s = 18.31 m/s
The vertical velocity as it left the cliff was v, where
v cos35° = 18.31
So, the vertical velocity was -v sin35° = -18.31 tan35° = -12.82 m/s
v(t) = -12.82 - 9.81*0.71 = -19.785 m/s
Now you have the vertical and horizontal components of the velocity upon impact.