The clubs would only collide, when the first one is on its falling path and while the second one is on its rising path.
Using S = ut + (1/2)gt^2. Noting that s = displacement from the point of release.
u = 15.
g= -10.
When the two clubs meet, let the first club be travelling for t seconds, the second club would be travelling for (t-2) seconds, since it was thrown 2 seconds later.
For first club, displacement s:
s = 15*t + (1/2) *(-10)t^2 = 15t-5t^2. ......(i)
For the second club, displacement s:
s = 15*(t-2) + (1/2) *(-10)(t-2)^2 = 15t-30 -5 (t-2)^2 = 15t-30 -5(t^2 -4t +4). Expanding the quadratic function.
s = 15t - 30 -5t^2 +20t - 20 = - 5t^2+35t -50
s= - 5t^2+35t -50 ...............................................(ii)
Equating equations (i) and (ii).
Therefore: 15t-5t^2 = - 5t^2+35t -50 Solving for t.
15t-5t^2 + 5t^2-35t = -50.
15t - 35t = -50
-20t = -50. Divide both sides by -20.
t = -50/-20 = 2.5 seconds.
Height above, using equation (i):
s=15t-5t^2, t = 2.5 s
s = 15*2.5 - 5*(2.5^2) = 37.5 - 31.25 = 6.25
Therefore height above = 6.25 m.
Velocity: v = u +gt. Note
For the first club. v = 15 + (-10)*2.5 = 15-25 = -10 m/s (Coming downwards).
For the second club, Note when they meet, at t=2.5s, the second club has only traveled (t-2) seconds. = (2.5-2) = 0.5 s
For the second club. v = 15 + (-10)*0.5 = 15-5 = 10 m/s (Going upwards).
That's it. Cheers.
