Question

A porpoise jumps straight up and crashes back into the water at 8.9 m/s. The drag and buoyancy forces of the water slow the porpoise down with an acceleration of -9.3 m/s2 as the porpoise finally slows to a stop. Find the depth the porpoise reaches.

Answer 1

$h=\frac{v_f^2-v_i^2}{2a},$ $v_f=0$ $\to$ $h=\frac{-v_i^2}{2a}=\frac{-8.9^2}{2\cdot(-9.3)}=4.26m$