# Projectile Motion

1. Dec 5, 2009

### whiteman

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
s = ut + 1/2 at2

3. 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 travelled 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.

2. Dec 5, 2009

### rl.bhat

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.

3. Dec 5, 2009

### whiteman

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?

4. Dec 5, 2009

### rl.bhat

In the expression of Ua and Ub sin(A-B) should be there instead of sin(A+B)