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A person on the board of a swimming pool throws a ball (ball1) at a speed of 8 m / s at an angle of 40 degrees above the horizontal. At the time of launch, the ball is 3 m above the water, and 10 m from the opposite side of the pool. From the opposite side of the pool, someone else throws another ball(ball2) at an angle of 50 degrees above the horizontal.

If the two are thrown at the same time: a) what should be speed of ball2 so that it touches ball1 in flight?

I just don't know where to start... it seems as though there is information missing! I have tried decomposing the speed of ball1 : Vxo = 8cos40=6.13m/s and Vyo = 8sin40 = 5.14m/s but I don't know where to go from there... I know I need to show the speed where x1=x2 and y1=y2, but how do I do that??

For projectiles we are given these equations :

x = (Vxo)(t)

Vy=(Vyo) - (g)(t)

y = (yo) + (Vyo)(t) - 0.5(g)(t)^2

Vy^2 = (Vyo)^2 - 2(g)(y-yo)

I have no idea how to solve this... I have been thinking about this one for at least 2 hours. Please help! Thank you so much in advance :)