We know ( v = \fracdsdt = 3t^2 ). Integrate:
Ground: ( s = 0 ). Use ( v^2 = v_0^2 + 2a(s - s_0) ): [ v^2 = 20^2 + 2(-9.81)(0 - 50) ] [ v^2 = 400 + 981 = 1381 ] [ v = -\sqrt1381 \quad (\textnegative because downward) ] [ \boxedv \approx -37.16 , \textm/s ] rectilinear motion problems and solutions mathalino
Since the particle moves to increasing ( s ) from rest at ( s=1 ), take positive root. We know ( v = \fracdsdt = 3t^2 )