Jamilla throws a stone horizontally off of a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0 m 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? Vf= 20m/s Vi= 0m/s a=9.8m/s2 Δd= 40m Δd=? Vf2=Vi2+2aΔd Vf=velocity final Vi= velocity initial a= acceleration Δd= displancement (rearrange the equation) Δd= (Vf2-Vi2) / 2a Δd= (20^2-o^2) / 2x9.8 Δd= 400/19.6 Δd= 20.4m ∴ the stone was 20.4m above the water surface when Jamilla threw it Not sure if i am doing this correctly? Much thanks in advance to those who help me out!