Problem: Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it? Related Equations: displacement = velocity x time displacement = (v1) (t) + (a) (t2) /2 t= square root of (2) (d) / a Solution: t= square root of (2) (d) /a t= square root of (2) (40.0m) /9.8m/s2 t= square root of 80m/ 9.8m/s2 t= square root of 8.16 t= 2.9s d= (v)(t) d= (20m/s) (2.9s) d= 58.0m t= time d= displacement a= acceleration Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation? Therefore the stone was 58.0m high above the ground when Jamilla threw it.