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Answer 1 · 2016-12-30T17:59:36

#sf(16.2color(white)(x)"m/s")#

MFDocs

See fig. (a)

First we find the vertical component of the motion using:

#sf(v^2=u^2+2as)#

This becomes:

#sf(v_y^2=0+2xxgxxh)#

#sf(v_y^2=0+2xx9.81xx6=117.72)#

#sf(v_y=sqrt(117.72)=10.85color(white)(x)"m/s")#

See fig (b).

The horizontal component of velocity is constant so we can find the resultant velocity #sf(V_(res))# using Pythagoras:

#sf(12^2+10.85^2=V_(res)^2)#

#:.##sf(V_(res)^2=261.72)#

#sf(V_(res)=sqrt(261.72)=16.2color(white)(x)"m/s")#

If you were asked to find the angle to the vertical then:

#sf(tantheta=12/10.85=1.105)#

From which #sf(theta=48^@)#