# Projectile Motion

## 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.

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 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.

rl.bhat
Homework Helper
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?

rl.bhat
Homework Helper
In the expression of Ua and Ub sin(A-B) should be there instead of sin(A+B)