
SUVAT equation help!!
Q. Two clay pidgeons are launched vertically upwards from exactly the same spot at 1 s intervals. Each clay pidgeon has an initial speed 30 m/s and acceleration 10 m/s downwards. How high above the ground do they collide?
I know i need to find out when both the pidgeons have the same height using a suvat, but im not sure how i would use an equation as i only have the acceleration and the initial speed.
Can anyone give me a hint, or even tell me the whole method. Thanks in advanced.

SUVAT?
I know Math is a language, but suvat equations?
So I searched Google.
Google referred me to Wikipedia.
Wikipedia says, "The SUVAT equations are five basic equations used to describe motion of a classical system under constant acceleration. They are named SUVAT equations after the five variables that they contain."
s Displacement. Units of m (meters, i.e distance and direction from start. It is a vector quantity).
u Initial velocity. Units of ms  1 (meters per second, i.e speed and direction. Is a vector quantity).
v Final velocity. Units of ms  1 (meters per second, i.e speed and direction. Is a vector quantity).
a Acceleration. Units of ms  2 (meters per second squared, i.e rate of change of speed, and direction. Is a vector quantity).
t Time. Units of s (seconds, i.e an amount of time. Is a scalar quantity).
v = u +at (1)
s = [(u +v)/2]*t (2)
And 3 more formulas (two for s and one for v^2)that can be easily derived from the two mentioned above.
I have memorized (1) and (2), in different forms, long time ago, and they are all I need to take care of these socalled SUVAT equations now. :)
Final velocity, Vf = Vo +at (1')
distance travelled, d = [(Vo +Vf)/2]*t (2')

Now to your question.
Q. Two clay pidgeons are launched vertically upwards from exactly the same spot at 1 s intervals. Each clay pidgeon has an initial speed 30 m/s and acceleration 10 m/s downwards. How high above the ground do they collide?
The two clay pigeons will collide when the first one is already going down while the second is still rising up at the same height above the ground.
s1 = s2 (i)
Given:
u1 = u2 = 30 m/sec upwards
a1 = a2 = 10 m/sec/sec downwards. so it is 10 m/sec/sec as we take upwards to be positive.
If we make the time the first one is fired as our reference, then
t1 = t seconds
t2 = (t1) seconds because t2 is less than t1.
We use the suvat equation, s = ut +a(t^2)/2:
s1 = 30t +(10)(t^2)/2
s2 = 30(t1) +(10)[(t1)^2]/2
s1 = s2
So,
30t +(10)(t^2)/2 = 30(t1) +(10)[(t1)^2]/2
30t 5t^2 = 30t 30 5[t^2 2t +1]
5t^2 = 30 5t^2 +10t 5
0 = 10t 35
t = 35/10 = 3.5 sec
Hence,
s1 = 30(3.5) 5(3.5)^2 = 43.75 m
s2 = 30(3.5 1) 5(3.5 1)^2 = 43.75 m
Therefore, the two clay pigeons will collide at 43.75 meters above the ground.