Solve #0.342costheta+hcos^2theta=16(D)^2/V_0^2#, if #V_0=60#, #D=80# and #h=2#?

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Answer 1 · 2017-04-09T16:31:57

There is no solution.

As #V_0=60#, #D=80# and #h=2#

#0.342costheta+hcos^2theta=16D^2/V_0^2# becomes

#0.342costheta+2cos^2theta=16xx6400/3600=256/9#

Hence this becomes

#18cos^2theta+3.078costheta-256=0#

and using quadratic formula

#costheta=(-3.078+-sqrt(3.078^2+4xx18xx256))/36#

= #(-3.078+-sqrt(9.4741+18432))/36#

= #(-3.078+-135.8)/36#

= #3.6868# or #-3.858#

As domain of #costheta# is #{-1,1}# and roots are beyond this range

There is no solution.