A pebble is released from rest at a certain height and falls freely, making an impact with a speed of 4 m/s at the floor. Next, the pebble is thrown down with an initial speed of 3 m/s from the same height. What is its speed just before it collides with  the floor?

Take downward direction as the positive direction. Since the pebble is released from rest, ${{\mathrm{v}}_{\mathrm{f}}}^2 = {{\mathrm{v}}^2}_{\mathrm{i}}+2\mathrm{a}∆\mathrm{y} \mathrm{becomes}$

${{\mathrm{v}}_{\mathrm{f}}}^2 = {( 4 \mathrm{m}/\mathrm{s})}^2 = 0^2+2\mathrm{gh}$

Next, when the pebble is thrown with speed 3.0 m/s from the same height h, we have

${{\mathrm{v}}^2}_{\mathrm{f}} = {(3 \mathrm{m}/\mathrm{s})}^2+2\mathrm{gh} = {(3 \mathrm{m}/\mathrm{s})}^2+{(4\mathrm{m}/\mathrm{s})}^2$

$\Rightarrow {\mathrm{v}}_{\mathrm{f}} = 5 \mathrm{m}/\mathrm{s}$