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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

$$\mathrm{suppose}\mathrm{that}\mathrm{two}\mathrm{rocks}\mathrm{are}\mathrm{thrown}\mathrm{from}$$$$\mathrm{thesame}\mathrm{point}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{moment}\mathrm{as}\mathrm{in}$$$$\mathrm{the}\mathrm{figure}\mathrm{above}.\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}$$$$\mathrm{them}\mathrm{as}\mathrm{a}\mathrm{function}\mathrm{of}\mathrm{time}.\mathrm{Assume}\mathrm{that}$$$$v_0\mathrm{and}{\theta }_0\mathrm{are}\mathrm{given}.$$

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

$$In{t}_{1}timeAparticlewillcross$$$$S_y=v_{0}sin\theta t_1-\frac{1}{2}{gt}_1^2\left(upward\right)$$$$Bparticlegoes{S}_y^{\prime }=v_{0}sin\theta t_1-\frac{1}{2}{gt}_1^2$$$$S_y-S_y^{\prime }=0$$$$In{t}_{1}Aparticlegoes\left(v_{0}cos{\theta }_0\right)t_{1}toward+xdirection$$$$Bparticlegoes\left(v_{0}cos{\theta }_0\right)t_{1}in-xdirection$$$$Relativedistance{S}_x=2v_{0}cos{\theta }_0{t}_1$$$$\sqrt{S_x^2+{(S_y-S_y^{\prime })}^2}=2v_{0}cos{\theta }_0{t}_1$$

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

$$\mathrm{thanks}$$

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

$$Ifvelocitiesarenotequal$$$$S_y=v_{1}sin\theta .t-\frac{1}{2}{gt}^2$$$$S_y^{\prime }=v_{2}sin\theta t-\frac{1}{2}{gt}^2$$$$(S_y-S_y^{\prime })=sin\theta (v_1-v_2)t$$$$S_x^{\prime }=(v_2+v_1)cos\theta .t$$$$\sqrt{S_x^{{}^{\prime }2}+{(S_y-S_y^{\prime })}^2}=t\sqrt{{(v_1-v_2)}^2-{cos}^2\theta ({(v_1-v_2)}^2-{(v_1+v_2)}^2)}$$$$=t\sqrt{{(v_1-v_2)}^2+4v_1{v}_2{cos}^2\theta }$$

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

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

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