Answer
0.4 s
Work Step by Step
Let's use the equation 2.9, $V^{2}=u^{2}+2aS$ to find its velocity when it is 14m above the ground.
$\downarrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V^{2}=0^{2}+2\times9.8\space m/s^{2}(53\space m -14\space m)$
$V^{2}=764.4\space m^{2}/s^{2}$
$V=27.7\space m/s$
Given that the block then falls the additional 12.0 m to the level of the man's head. So Let's apply equation 2.8, $S=ut+\frac{1}{2}at^{2}$ to find the time taken,
$\downarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$12=27.7t+\frac{1}{2}9.8t^{2}$
$4.9t^{2}+27.7t-12=0$
From the quadratic formula, we obtain
$$t=\frac{-27.7\pm \sqrt {27.7^{2}-4(4.9)(-12)}}{2\times 4.9}=0.4\space s\space or\space -6.1\space s$$
Time isn't a minus value so we neglect the negative solution. Then the time it takes for the block to reach the level of the man is = 0.4 s
