If someone would just help me with setting up the right equations for the proble

Question

If someone would just help me with setting up the right equations for the problem that would be awesome!

Explanation / Answer

U= 28m/sec theta = 37.8 degree Uhor= 28cos(37.8)=22.12m/sec Uver=28sin(37.8) =17.16m/sec There is acceleretion of -g in vertical direction only At highest point Vver=0 Vver=Uver-g*t =>time to reach highest point = 17.16/9.8=1.751 sec It will take twicw of this time to fall back on ground =2*1.751=3.5 sec Horizontal distance covered in this time =22.12*3.5= 77.465m a) 77.465 m b)Hmax= Uver*t' - 0.5*g*t'^2 [where t' is time to reach hmax=1.75 sec] =>Hmax= 15.02 m c)At highest point Vver = 0 , Vhor = 22.12 m/sec V=22.12 i m/sec d)Every where Aver =-g , Ahor=0 At highest point A= -g j m/sec^2 [ver stands for vertical and hor for horizontal]

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