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An-aircraft-flying-horizontally-100m-above-the-ground-and-at-720km-h-drops-a-bomb-on-a-target-on-the-ground-Determine-the-acute-angle-between-the-vertical-and-the-line-joining-the-aircraft-and-target-




Question Number 47333 by Necxx last updated on 08/Nov/18
An aircraft flying horizontally  100m above the ground and at 720km/h  drops a bomb on a target on the  ground.Determine the acute angle  between the vertical and the line  joining the aircraft and target at  the instance when the bomb is  released.(g=10ms^(−2) )
$${An}\:{aircraft}\:{flying}\:{horizontally} \\ $$$$\mathrm{100}{m}\:{above}\:{the}\:{ground}\:{and}\:{at}\:\mathrm{720}{km}/{h} \\ $$$${drops}\:{a}\:{bomb}\:{on}\:{a}\:{target}\:{on}\:{the} \\ $$$${ground}.{Determine}\:{the}\:{acute}\:{angle} \\ $$$${between}\:{the}\:{vertical}\:{and}\:{the}\:{line} \\ $$$${joining}\:{the}\:{aircraft}\:{and}\:{target}\:{at} \\ $$$${the}\:{instance}\:{when}\:{the}\:{bomb}\:{is} \\ $$$${released}.\left({g}=\mathrm{10}{ms}^{−\mathrm{2}} \right) \\ $$$$ \\ $$
Answered by MrW3 last updated on 08/Nov/18
v=720km/h=200m/s  x=vt  y=((gt^2 )/2)⇒t=(√((2y)/g))  ⇒x=v(√((2y)/g))  tan θ=(x/y)=v(√(2/(gy)))=200(√(2/(10×100)))=((20)/( (√5)))  ⇒θ=tan^(−1) ((20)/( (√5)))=83.6°
$${v}=\mathrm{720}{km}/{h}=\mathrm{200}{m}/{s} \\ $$$${x}={vt} \\ $$$${y}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{t}=\sqrt{\frac{\mathrm{2}{y}}{{g}}} \\ $$$$\Rightarrow{x}={v}\sqrt{\frac{\mathrm{2}{y}}{{g}}} \\ $$$$\mathrm{tan}\:\theta=\frac{{x}}{{y}}={v}\sqrt{\frac{\mathrm{2}}{{gy}}}=\mathrm{200}\sqrt{\frac{\mathrm{2}}{\mathrm{10}×\mathrm{100}}}=\frac{\mathrm{20}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{20}}{\:\sqrt{\mathrm{5}}}=\mathrm{83}.\mathrm{6}° \\ $$
Commented by Necxx last updated on 08/Nov/18
thank you sir. This was what I also  got but the answer says 70.5 in    the text
$${thank}\:{you}\:{sir}.\:{This}\:{was}\:{what}\:{I}\:{also} \\ $$$${got}\:{but}\:{the}\:{answer}\:{says}\:\mathrm{70}.\mathrm{5}\:{in} \\ $$$$ \\ $$$${the}\:{text} \\ $$
Commented by MrW3 last updated on 08/Nov/18
then the answer in book is wrong.
$${then}\:{the}\:{answer}\:{in}\:{book}\:{is}\:{wrong}. \\ $$
Answered by ajfour last updated on 08/Nov/18
h=(1/2)gt^2   D=vt  h=(g/2)((D/v))^2   ⇒   D= v(√((2h)/g))   tan θ = (D/h) = ((v(√2))/( (√(gh)))) = ((200(√2))/( (√(1000)))) = 4(√5)    θ = tan^(−1) (4(√5)) .
$${h}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${D}={vt} \\ $$$${h}=\frac{{g}}{\mathrm{2}}\left(\frac{{D}}{{v}}\right)^{\mathrm{2}} \:\:\Rightarrow\:\:\:{D}=\:{v}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\: \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{{D}}{{h}}\:=\:\frac{{v}\sqrt{\mathrm{2}}}{\:\sqrt{{gh}}}\:=\:\frac{\mathrm{200}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1000}}}\:=\:\mathrm{4}\sqrt{\mathrm{5}}\: \\ $$$$\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{4}\sqrt{\mathrm{5}}\right)\:. \\ $$

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