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Answer 1 · 2015-04-10T11:32:52

#32.1" m/s "##[82^o" below horizontal"]#

First divide the initial velocity vector into horizontal components and vertical components.

The horizontal component will be

#v_x=vcostheta = 7.0cos53# m/s

I am assuming we are neglecting air resistance so this value will remain constant. #v_(fx) = v_(ix) = v_x =4.21# m/s

The vertical component will be

#v_y=vsintheta = 7.0sin53# m/s [up]

#v_(fy)# the final velocity in the y-direction can be determined by

#(v_(fy))^2=(v_(iy))^2 + 2gDeltay# . . . . substitution gives

#v_(fy)=sqrt((7.0sin53)^2 + 2(-9.81)(-50))=31.82# m/s

Now combine the two components to get the final velocity
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#v_f=sqrt(v_(fx)^2 + v_(fy)^2) = 32.1# m/s

#theta = tan^(-1)((v_"fy")/(v_"fx"))=tan^(-1)((31.82)/(4.21))=82^o#