Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 61: 28

$a). 9.8 m/s^{2}$ $b). 4.69s$

$v=u+at$ a). after the object is released, its acceleration with respect to the ground is $9.8 m/s^{2}$ in vertically downward direction. b). $v^{2}=u^{2}+2aS$ $15^{2}=0+2(3)S$ $S=\frac{225}{6}=37.5m$ Now.$S=ut+\frac{1}{2}at^{2}$ or, $37.5=-15t+0.5\times 9.8\times t^{2}$ $t=4.69s$