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Question Number 114475 by Rio Michael last updated on 19/Sep/20

Commented by Rio Michael last updated on 19/Sep/20

suppose that two rocks are thrown from thesame point at the same moment as in the figure above. Find the distance between them as a function of time. Assume that v_0 and θ_0 are given.

$$\mathrm{suppose}\:\mathrm{that}\:\mathrm{two}\:\mathrm{rocks}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{from} \\ $$$$\mathrm{thesame}\:\mathrm{point}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{moment}\:\mathrm{as}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{figure}\:\mathrm{above}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{them}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}.\:\mathrm{Assume}\:\mathrm{that} \\ $$$${v}_{\mathrm{0}} \:\mathrm{and}\:\theta_{\mathrm{0}} \:\mathrm{are}\:\mathrm{given}. \\ $$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

In t_1 time A particle will cross S_y =v_0 sinθt_1 −(1/2)gt_1 ^2 (upward) B particle goes S_y ′=v_0 sinθt_1 −(1/2)gt_1 ^2 S_y −S_y ′=0 In t_1 A particle goes (v_0 cosθ_0 )t_1 toward +x direction B particle goes (v_0 cosθ_0 )t_1 in −x direction Relative distance S_x =2v_0 cosθ_0 t_1 (√(S_x ^2 +(S_y −S_y ′)^2 )) =2v_0 cosθ_0 t_1

$${In}\:{t}_{\mathrm{1}} \:{time}\:{A}\:{particle}\:{will}\:{cross}\: \\ $$$${S}_{{y}} ={v}_{\mathrm{0}} {sin}\theta{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\left({upward}\right) \\ $$$${B}\:{particle}\:{goes}\:{S}_{{y}} '={v}_{\mathrm{0}} {sin}\theta{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${S}_{{y}} −{S}_{{y}} '=\mathrm{0} \\ $$$${In}\:{t}_{\mathrm{1}} \:{A}\:{particle}\:{goes}\:\left({v}_{\mathrm{0}} {cos}\theta_{\mathrm{0}} \right){t}_{\mathrm{1}} \:\:{toward}\:+{x}\:{direction} \\ $$$${B}\:{particle}\:{goes}\:\left({v}_{\mathrm{0}} {cos}\theta_{\mathrm{0}} \right){t}_{\mathrm{1}} \:{in}\:−{x}\:{direction} \\ $$$${Relative}\:{distance}\:{S}_{{x}} =\mathrm{2}{v}_{\mathrm{0}} {cos}\theta_{\mathrm{0}} {t}_{\mathrm{1}} \\ $$$$\sqrt{{S}_{{x}} ^{\mathrm{2}} +\left({S}_{{y}} −{S}_{{y}} '\right)^{\mathrm{2}} }\:\:=\mathrm{2}{v}_{\mathrm{0}} {cos}\theta_{\mathrm{0}} {t}_{\mathrm{1}} \\ $$

Commented by Rio Michael last updated on 19/Sep/20

thanks

$$\mathrm{thanks} \\ $$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

If velocities are not equal S_y =v_1 sinθ.t−(1/2)gt^2 S′_y =v_2 sinθt−(1/2)gt^2 (S_y −S_y ′)=sinθ(v_1 −v_2 )t S′_x =(v_2 +v_1 )cosθ.t (√(S_x ^(′2) +(S_y −S_y ′)^2 )) =t(√((v_1 −v_2 )^2 −cos^2 θ((v_1 −v_2 )^2 −(v_1 +v_2 )^2 ))) =t(√((v_1 −v_2 )^2 +4v_1 v_2 cos^2 θ))

$${If}\:{velocities}\:{are}\:{not}\:{equal} \\ $$$${S}_{{y}} ={v}_{\mathrm{1}} {sin}\theta.{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${S}'_{{y}} ={v}_{\mathrm{2}} {sin}\theta{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\left({S}_{{y}} −{S}_{{y}} '\right)={sin}\theta\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right){t} \\ $$$${S}'_{{x}} =\left({v}_{\mathrm{2}} +{v}_{\mathrm{1}} \right){cos}\theta.{t} \\ $$$$\sqrt{{S}_{{x}} ^{'\mathrm{2}} +\left({S}_{{y}} −{S}_{{y}} '\right)^{\mathrm{2}} }\:={t}\sqrt{\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right)^{\mathrm{2}} −{cos}^{\mathrm{2}} \theta\left(\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({v}_{\mathrm{1}} +{v}_{\mathrm{2}} \right)^{\mathrm{2}} \right)} \\ $$$$={t}\sqrt{\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{v}_{\mathrm{1}} {v}_{\mathrm{2}} {cos}^{\mathrm{2}} \theta} \\ $$

Commented by Rio Michael last updated on 19/Sep/20

thanks for the extra

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{extra} \\ $$

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