Answer to Question #112058 in Mechanics | Relativity for Sam

Question

A human cannon ball to be ejected at a scarily fast velocity of 11 ms-1 at an angle of 43˚ to the horizontal. The plan is that the ‘volunteer’, Susie, will land in a large (deep) paddling pool, which needs to be strategically placed on the horizontal playing field. Ignoring any air resistance and using g = 9.8 ms-2, calculate: (3.1)


(a) time taken to reach her maximum height above the ground



(b) total time spent in the air


(c) horizontal distance to where the paddling pool should be placed


(d) Susie’s velocity as she (hopefully) hits the water.


(e) If the pool has to been placed on ground that is 2.5 m lower than the launch point, calculate her new splash-down velocity.

Answer 1

$a)v_y=v_{y0}-gt \\ 0=v\sin 43^{\circ}-9.8t\\t=\frac{11\sin43^{\circ}}{9.8}=\frac{11\times0.68{}}{9.8}=0.763s\\b)h=y_0+v_0\times\sin{\alpha}\times t -\frac{g\times t^2}{2}\\0=11\times\sin43^{\circ}\times t -\frac{9.8\times t^2}{2}\\11\times\sin43^{\circ}\times t -\frac{9.8\times t^2}{2}=0;\\t\times(\frac{9.8\times t}{2}-11\times0.68)=0;\\t=0;\\t=\frac{2\times11\times0.68}{9.8}=1.53s\\c)x_{max}=t(b)\times v_0\times\cos43^{\circ}=\\=1.53\times11\times0.73=12.29m\\d)v_{x0}=v_x;d)v_{y}=v_{y0}-g\times t(b)=\\=11\times\sin43^{\circ}-9.8\times1.53=\\=7.48-11.994=-7.48\frac{m}{s}\\v_0=v=11\frac{m}{s}\\e)y=v_{y0}\times t+\frac{g\times t^2}{2};\\2.5=11\sin43^{\circ}\times t+\frac{9.8\times t^2}{2};\\4.9\times t^2+7.48\times t-2.5=0;\\D=55.95+49=104.95;\\t=\frac{-7.48+10.24}{9.8}=0.28s;\\v_{x0}=const=v\times cos{43^{\circ}}=8.03\frac{m}{s};\\v_y=v\times sin43^{\circ}+g\times t=7.48+2.74=10.22\frac{m}{s};\\v^2=8.03^2+10.22^2=64.48+104.45=169;\\v=13\frac{m}{s}$