Question

Question #0c86a

Answer 1

Discussed below

Explanation:

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`