Answer
(a) Minimum required speed of the rock, $u_{min}=13.3 \space m/s$
(b) Rock lands on the ground at distance $D=3.75 \space m$ from the fence.
Work Step by Step
(a) Using equation of trajectory,
$y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}$
At $x=14\space m$, minimum value of y is, $ y=5.0-1.6=3.4\space m$ .
So, $3.4=14tan56^0-\frac{9.8\times14^2}{2u^2cos^256^0}\Rightarrow u_{min}=\sqrt \frac{9.8\times14^2}{2u^2cos^256^0(14tan56^0-3.4)}=13.3 \space m/s$
(b) When ball falls on the ground, $y=-1.6\space m$ with respect to point of projection,
So using equation of trajectory,
$y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta} \Rightarrow -1.6=xtan56^0-\frac{9.8\times x^2}{2\times13.3^2cos^256^0}$
Solving this quadratic equation in x, we positive solution of x as, $x=17.75\space m$.
So, rock lands on the ground at distance $D=17.75-14=3.75 \space m$ from the fence.
