1. The problem statement, all variables and given/known data A kid stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.6 m/s. The cliff is 48.8 m above the ocean. How long after being released does the stone strike the water? With what speed and angle of impact does it hit the water? 2. Relevant equations y = V_oy * t + 1/2 * g * t² x = v_ox * t 3. The attempt at a solution Well to find the time I just used the first equation...plugged in -48.8 m for y, initial velocity in the y direction is 0.....g is going to be -9.8 here...and finally t² So t² = -48.8 / -4.9 ...and t ≈3.2 seconds Now for the distance I used the second equation...plugged 16.6 in for V_ox and the 3.2 for t which came out to ≈ 52.4m But the angle part is confusing me...how do I figure out the angle? I know that i would normally use the y components and x components....but I dont have an angle because the stone was thrown horizontally.... am I supposed to use the cliff 48.8sin(90) for the y component (no x component there)....and use 16.6cos(0) for the x component? (no y component there)....So would the answer be inverse tangent (48.8/16.6) = ≈71°?