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.

Solve 0.342costheta+hcos^2theta=16(D)^2/V_0^20.342costheta+hcos^2theta=16(D)^2/V_0^2, if V_0=60V_0=60, D=80D=80 and h=2h=2?