Finding Speed of Projection for Two Balls Collision

AI Thread Summary
The discussion centers on calculating the speed of projection for two balls that collide after being projected at different angles and times. The equations of motion in both x and y directions are established, with the key condition that the distances traveled by both balls are equal at the time of collision. The first ball is in the air for 2 seconds, while the second is in the air for 1 second. The user derived expressions for the speeds of both balls but questioned the correctness of their formulation regarding the sine functions involved. The final consensus suggests that the expressions for the speeds should include sin(A-B) instead of sin(A+B).
whiteman
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Homework Statement


A ball was projected at an angle A to the horizontal. One second later another ball was projected from the same point at an angle B to the horizontal. One second after the second ball was released, the two balls collided. Find the speed of projection for the two balls.


Homework Equations


s = ut + 1/2 at2


The Attempt at a Solution


In x-direction:
sA = ua cosA t
sB = ub cosB (t-1)

In y-direction:
sA = ua sinA t -1/2 gt2
sB = ub sinB (t-1) -1/2 g(t-1)2

When they collide the distances they traveled in the x and y directions are equal to each other at t=2. I tried to solve for ua and ub but got stuck.
 
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Hi whiteman, welcome to PF.
When the balls collide, ball A is in the air for 2 s and ball B is in the air for 1 s.
For them x distance is the same. So
2*Ua*cosA = Ub*cosB -----(1)
For y
2*Ua*sinA - 1/2*g*(2)^2 = Ub*sinB - 1/2*g -------(2)
From eq.(1), find the expression for Ub and substitute it in equation (2) and solve for Ua.
 
Hi, thanks for the welcome.
Is Ua and Ub meant to expressed in terms of cos/sin of A/B or are they just numbers?
I got Ua = 3gcosB/4sin(A+B) and Ub = 3gcosA/2sin(A+B).
Is this right or am I going wrong?
 
In the expression of Ua and Ub sin(A-B) should be there instead of sin(A+B)
 
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