Answer
(a) Speed with which ball strikes the building, $u \approx 16.6 m/s$
(b) The at which ball strikes the building, $ \theta \approx 40.5^{\circ}$ (below horizontal)
Work Step by Step
(a) Using equation of trajectory,
$y=xtan\theta- \frac{gx^{2}}{2u^{2}cos^{2}\theta}$
$8=18tan60^{\circ}- \frac{9.8\times 18^{2}}{2u^{2}cos^{2}60^{\circ}}$
$8=18\sqrt 3-\frac{9.8\times 18^{2}}{2u^{2}(\frac{1}{4})}$
On solving, we get $u \approx 16.6 m/s$
(b) When ball is about to strike the building,
Along x-axis : $v_{x}=u_{x}+a_{x}t=ucos60^{\circ}+0\times t=\frac {u}{2}$
Along y-axis : $v_{y}^{2}=u_{y}^{2}+2a_{y}s_{y} \approx 8.3 m/s$
$v_{y}=\sqrt{(usin60^{\circ})^{2}+2\times (-9.8)(8)}\approx 7.1 m/s$ (It is directed downwards)
So net speed of the ball, $v=\sqrt{v_{x}^{2}+v_{y}^{2}}\approx 10.9 m/s$
Also, $tan\theta=\frac{v_{y}}{v_{x}}=\frac{7.1}{8.3}\approx 0.85$
So,$ \theta \approx 40.5^{\circ}$ (below horizontal)
