Answer
(a) Bolt will hit the floor after $t=0.78 \space s$.
(b) Speed of the bolt w.r.t. observer in the lift, $v_{bl} =7.64 \space m/s$.
(c) Speed of the bolt w.r.t. observer on the ground, $v_b=5.14 \space m/s$.
(d) Displacement of the bolt w.r.t. observer on the ground, $s_b=1.03 \space m$ (downward direction).
Work Step by Step
Motion of the bolt relative to lift (assuming downward direction to be positive),
Initial relative velocity $u_{bl}=0$, relative acceleration $a_{bl}=a_{b}-a_{l}=g-0=g$, relative displacement $s_{bl}=3 \space m$.
(a) Using equation $s=ut+\frac{1}{2}at^2$ for bolt w.r.t. lift,
$-3=0\times t+\frac{1}{2}(9.8)t\Rightarrow t=\sqrt\frac{2\times 3}{9.8}=0.78 \space s$
(b) Speed of the bolt w.r.t. observer in the lift,
$v_{bl}=u_{bl}+a_{bl}t\Rightarrow v_{bl}=0+9.8\times 0.78=7.64 \space m/s$
(c) Speed of the bolt w.r.t. observer on the ground,
$v_b=u_b+a_bt\Rightarrow v_{bl}=-2.50+9.8\times 0.78=5.14 \space m/s$
(d) Displacement of the bolt w.r.t. observer on the ground,
$s_b=u_bt+\frac{1}{2}a_bt^2\Rightarrow s_b=-2.50\times 0.78+\frac{1}{2}(9.8)(0.78)^2=1.03 \space m$ (downward direction)
