"∆y = h? V_oy = sin60(31) = 31√3/2 V_fy = 0 (pretty sure its safe to assume that it barely reached the cliff, so Vfy = 0) t = 3.7 s a = -9.8 m/s^2 ∆y = V_o * t + (1/2)at^2 ∆y = (31√3/2)(3.7) + (1/2)(-9.8)(3.7^2) ∆y = h = 32.25 m Now to find impact speed i just said V_fy = 0, so there must be some V_fx. ∆x? V_ox = 31cos60 = 31/2 V_fx = ? t = 3.7 s (same time as for y) a = 0 (In free fall there is only acceleration in the y direction, no x) V = V_o + at but since a = 0, there is nothing messing up the x directions trajectory since we are ignoring air resistance. So it will have the same velocity of: Vo = Vf = 31cos60 = 31/2 = 15.5 m/s"