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Question-27044

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Question Number 27044 by Tinkutara last updated on 01/Jan/18
Answered by mrW1 last updated on 02/Jan/18
α=30° x=u cos θ t y=u sin θ t −((gt^2 )/2) by landing: y=x tan α=((x sin α)/(cos α)) t_1 =3.5 s ⇒ut_1 sin θ−((gt_1 ^2 )/2)=((sin α)/(cos α))×ut_1 cos θ ⇒sin θ−((gt_1 )/(2u))=((sin α)/(cos α)) cos θ ⇒cos α sin θ−((gt_1 cos α)/(2u))=sin α cos θ ⇒cos α sin θ−sin α cos θ=((gt_1 cos α)/(2u)) ⇒sin (θ− α)=((gt_1 cos α)/(2u)) ⇒θ= α+sin^(−1) (((gt_1 cos α)/(2u))) ⇒θ= 30+sin^(−1) (((10×3.5×cos 30°)/(2×53)))=46.62° at t=2 s: u_x (t)=u cos θ u_y (t)=u sin θ−gt u(t)=(√((u cos θ)^2 +(u sin θ−gt)^2 )) u(t)=(√(u^2 +gt(gt−2u sin θ))) u(t)=(√(53^2 +10×2(10×2−2×53×sin 46.62°)))=40.84 m/s Δu=53−40.84=12.16 m/s
$$\alpha=\mathrm{30}° \\ $$$${x}={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${y}={u}\:\mathrm{sin}\:\theta\:{t}\:−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${by}\:{landing}: \\ $$$${y}={x}\:\mathrm{tan}\:\alpha=\frac{{x}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha} \\ $$$${t}_{\mathrm{1}} =\mathrm{3}.\mathrm{5}\:{s} \\ $$$$\Rightarrow{ut}_{\mathrm{1}} \:\mathrm{sin}\:\theta−\frac{{gt}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}×{ut}_{\mathrm{1}} \:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{sin}\:\theta−\frac{{gt}_{\mathrm{1}} }{\mathrm{2}{u}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta−\frac{{gt}_{\mathrm{1}} \:\mathrm{cos}\:\alpha}{\mathrm{2}{u}}=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta−\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta=\frac{{gt}_{\mathrm{1}} \:\mathrm{cos}\:\alpha}{\mathrm{2}{u}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\theta−\:\alpha\right)=\frac{{gt}_{\mathrm{1}} \:\mathrm{cos}\:\alpha}{\mathrm{2}{u}} \\ $$$$\Rightarrow\theta=\:\alpha+\mathrm{sin}^{−\mathrm{1}} \left(\frac{{gt}_{\mathrm{1}} \:\mathrm{cos}\:\alpha}{\mathrm{2}{u}}\right) \\ $$$$\Rightarrow\theta=\:\mathrm{30}+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{10}×\mathrm{3}.\mathrm{5}×\mathrm{cos}\:\mathrm{30}°}{\mathrm{2}×\mathrm{53}}\right)=\mathrm{46}.\mathrm{62}° \\ $$$$ \\ $$$${at}\:{t}=\mathrm{2}\:{s}: \\ $$$${u}_{{x}} \left({t}\right)={u}\:\mathrm{cos}\:\theta \\ $$$${u}_{{y}} \left({t}\right)={u}\:\mathrm{sin}\:\theta−{gt} \\ $$$${u}\left({t}\right)=\sqrt{\left({u}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({u}\:\mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} } \\ $$$${u}\left({t}\right)=\sqrt{{u}^{\mathrm{2}} +{gt}\left({gt}−\mathrm{2}{u}\:\mathrm{sin}\:\theta\right)} \\ $$$${u}\left({t}\right)=\sqrt{\mathrm{53}^{\mathrm{2}} +\mathrm{10}×\mathrm{2}\left(\mathrm{10}×\mathrm{2}−\mathrm{2}×\mathrm{53}×\mathrm{sin}\:\mathrm{46}.\mathrm{62}°\right)}=\mathrm{40}.\mathrm{84}\:{m}/{s} \\ $$$$\Delta{u}=\mathrm{53}−\mathrm{40}.\mathrm{84}=\mathrm{12}.\mathrm{16}\:{m}/{s} \\ $$
Commented by mrW1 last updated on 03/Jan/18
It can also be understood like this: Δu_x =0 Δu_y =−gt=−10×2=−20 m/s ⇒∣Δu∣=(√(0^2 +(−20)^2 ))=20 m/s
$${It}\:{can}\:{also}\:{be}\:{understood}\:{like}\:{this}: \\ $$$$\Delta{u}_{{x}} =\mathrm{0} \\ $$$$\Delta{u}_{{y}} =−{gt}=−\mathrm{10}×\mathrm{2}=−\mathrm{20}\:{m}/{s} \\ $$$$\Rightarrow\mid\Delta{u}\mid=\sqrt{\mathrm{0}^{\mathrm{2}} +\left(−\mathrm{20}\right)^{\mathrm{2}} }=\mathrm{20}\:{m}/{s} \\ $$
Commented by Tinkutara last updated on 03/Jan/18
Yes answer given is 20 m/s. Thank you very much Sir! I got the answer.

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