Projectile Motion

  • Thread starter whiteman
  • Start date
  • #1
9
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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 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.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
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
9
0
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
rl.bhat
Homework Helper
4,433
8
In the expression of Ua and Ub sin(A-B) should be there instead of sin(A+B)
 

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