A ball is tossed straight up with an initial velocity of 48m/s. Exactly how fast is it going after 3.0 seconds

Answer 1

v(t) = v_i + at

v_i = 48 m/s

a = -9.8 m/s²

v(t) = 48 m/s + (-9.8 m/s²)(3.0 s) = 18.6 m/s

Note how the initial velocity and acceleration are in opposite directions.