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Question Number 46629 by Necxx last updated on 29/Oct/18

An object is projected from a height of 80m above the ground with a velocity of 40m/s at an angle of 30 degree to the horizontal.What is the tume of flight?

$${An}\:{object}\:{is}\:{projected}\:{from}\:{a} \\ $$$${height}\:{of}\:\mathrm{80}{m}\:{above}\:{the}\:{ground} \\ $$$${with}\:{a}\:{velocity}\:{of}\:\mathrm{40}{m}/{s}\:{at}\:{an} \\ $$$${angle}\:{of}\:\mathrm{30}\:{degree}\:{to}\:{the} \\ $$$${horizontal}.{What}\:{is}\:{the}\:{tume}\:{of} \\ $$$${flight}? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

−h=(usinθ)t−(1/2)gt^2 here displacement=−80 gravitational accelaration down ward so −ve −80=40×(1/2)×t−5t^2 −80=20t−5t^2 −16=4t−t^2 t^2 −4t−16=0 t=((4±(√(16+64)))/2) t=((4±4(√5))/2) so the answer is t=((4+4(√5))/2)=2+2(√5) second pls check t=2(√5) +2=6.47sec if we take g=9.8 then...eqn is −80=(1/2)×40t−(1/2)×9.8t^2 9.8t^2 −40t−160=0 t=((40±(√(1600+4×9.8×160)) )/(2×9.8)) =((40±(√(1600+64×98)) )/(2×9.8)) =((40±(√(7872)))/(2×9.8))=((40±88.72)/(2×9.8)) t=((128.72)/(2×9.8))=6.57

$$−{h}=\left({usin}\theta\right){t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\:\:\:\:{here}\:{displacement}=−\mathrm{80} \\ $$$${gravitational}\:{accelaration}\:{down}\:{ward}\:{so}\:−{ve} \\ $$$$−\mathrm{80}=\mathrm{40}×\frac{\mathrm{1}}{\mathrm{2}}×{t}−\mathrm{5}{t}^{\mathrm{2}} \\ $$$$−\mathrm{80}=\mathrm{20}{t}−\mathrm{5}{t}^{\mathrm{2}} \\ $$$$−\mathrm{16}=\mathrm{4}{t}−{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{16}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{64}}}{\mathrm{2}} \\ $$$${t}=\frac{\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${so}\:{the}\:{answer}\:{is}\:{t}=\frac{\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}\:{second} \\ $$$${pls}\:{check} \\ $$$${t}=\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}=\mathrm{6}.\mathrm{47}{sec} \\ $$$${if}\:{we}\:{take}\:{g}=\mathrm{9}.\mathrm{8} \\ $$$${then}...{eqn}\:{is} \\ $$$$−\mathrm{80}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{40}{t}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}.\mathrm{8}{t}^{\mathrm{2}} \\ $$$$\mathrm{9}.\mathrm{8}{t}^{\mathrm{2}} −\mathrm{40}{t}−\mathrm{160}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{40}\pm\sqrt{\mathrm{1600}+\mathrm{4}×\mathrm{9}.\mathrm{8}×\mathrm{160}}\:}{\mathrm{2}×\mathrm{9}.\mathrm{8}} \\ $$$$\:\:=\frac{\mathrm{40}\pm\sqrt{\mathrm{1600}+\mathrm{64}×\mathrm{98}}\:}{\mathrm{2}×\mathrm{9}.\mathrm{8}} \\ $$$$=\frac{\mathrm{40}\pm\sqrt{\mathrm{7872}}}{\mathrm{2}×\mathrm{9}.\mathrm{8}}=\frac{\mathrm{40}\pm\mathrm{88}.\mathrm{72}}{\mathrm{2}×\mathrm{9}.\mathrm{8}} \\ $$$${t}=\frac{\mathrm{128}.\mathrm{72}}{\mathrm{2}×\mathrm{9}.\mathrm{8}}=\mathrm{6}.\mathrm{57} \\ $$

Commented by Necxx last updated on 29/Oct/18

the options are a)2 b)4 c)8 d)10

$$\left.{t}\left.{h}\left.{e}\left.\:{options}\:{are}\:{a}\right)\mathrm{2}\:{b}\right)\mathrm{4}\:{c}\right)\mathrm{8}\:{d}\right)\mathrm{10} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

Commented by Necxx last updated on 29/Oct/18

thank you so much. this was also what i got but i was confused at not seeing any related option. I believe you prof.

$${thank}\:{you}\:{so}\:{much}.\:{this}\:{was}\:{also} \\ $$$${what}\:{i}\:{got}\:{but}\:{i}\:{was}\:{confused}\:{at} \\ $$$${not}\:{seeing}\:{any}\:{related}\:{option}. \\ $$$${I}\:{believe}\:{you}\:{prof}. \\ $$

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