Ajuggler throws one club vertically upwards at 15ms^-1 and 2 seconds later throws a second club vertically up at the same speed. find the height above the start that the two clubs are at the same height and the speeds that they are moving when the two clubs collide. take a= 10
s= ?
u= 15ms^-1
v= ?
a= -10
t=

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.

It has a slope of 1 nd a y intercept of -2

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