Question #0c86a

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Answer 1 · 2016-06-21T09:02:48

Discussed below

Given

$h->"The height from which the canon ball was fired"=60m$

$u->"The vlocity of projection"=25m/s$

$alpha -> "Angle of projection"=53^@$

$"And "sin53^@=0.8 and cos53^@=0.6$

Now

$"Horizontal component of vel.of projection " ucos53^@=25*0.6=15m/s$

$"Vertical component of vel.of projection " usin53^@=25*0.8=20m/s$

At the time when it reaches at maximum height H from its point of projection its vertical component becomes zero, So considering $g=ms^-2$ we get

$0^2=20^2-2*10*H=>H=20m$

Hence at its maximum height the total height fron ground
$H_"total"=60+20=80m$

Now if its time of flight be t s
then it will undergo a vertical displacement downward during this time. So
-
$-60=20*t-1/2*20*t^2$

$=>t^2-2t-6=0$

$=>t=(2+sqrt(4-4*(-6)*1))/2$

$=>t=(sqrt7+1)s$

Net horizontal displacement during this time of flight

$=ucosalpha*t=15(sqrt7+1)m$