Answer:
Ok, remember that the only force acting on the objects will be the acceleration force.
First, let's describe the motion of the rock.
The acceleration will be:
Ar(t) = -9.8m/s^2
(where the negative sign is because the gravity pulls the rock downwards).
If we want to obtain the velocity of the rock, we should integrate over time and get:
Vr(t) = (-9.8m/s^2)*t + v0
Where v0 is the initial velocity, and we know that it is:
v0 = 12.0m/s
Then:
Vr(t) = (-9.8m/s^)*t + 12m/s.
And the position can be obtained if we integrate again.
Pr(t) = (-4.9m/s^)*t^2 + 12m/s*t + p0
Where p0 is the initial position.
Now let's do the same for the ball, but the ball is thrown one second after, so we will work with the variable (t - 1s)
Ab(t) = -9.8m/s^2
Vb(t) = (-9.8m/s^2)*(t - 1s) + 18m/s
Pb(t) = (-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0
As the ball is thrown along the same path than the rock, the initial position is the same in both cases.
Now the two objects will impact each other when:
Pb(t) = Pr(t)
(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0 = (-4.9m/s^)*t^2 + 12m/s*t + p0
Let's solve this for t.
(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) = (-4.9m/s^)*t^2 + 12m/s*t
(-4.9m/s^2)*(t^2 -2*t*1s + 1s^2) + 18m/s*(t - 1s) = (-4.9m/s^)*t^2 + 12m/s*t
(-4.9m/s^2)*(-2*t*1s + 1s^2) + 18m/s*t - 18m = 12m/s*t
-4.9m + 9.8m/s*t + 18m/s*t - 18m = 12m/s*t
-22.9m = t*(12m/s - 9.8m/s - 18m/s)
22.9m = t*16.8m/s
(22.9/16.8)s = 1.363 s
So the ball and the rock will strike each other at t = 1.363 seconds.
