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

Commented byRio 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.

Commented byDwaipayan 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

Commented byRio Michael last updated on 19/Sep/20

thanks

Commented byDwaipayan 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))

Commented byRio Michael last updated on 19/Sep/20

thanks for the extra