Two children are playing on a 147-m-tall bridge. One child drops a rock (initial velocity zero) at

t = 0.
The other waits 1.2 s and then throws a rock downward with an initial speed
v0.
If the two rocks hit the ground at the same time, what is
v0?

To find the value of v0 (the initial speed of the rock thrown downward), we need to consider the motion of both rocks and determine the time it takes for each rock to hit the ground. We know that the height of the bridge is 147 m, and we want both rocks to hit the ground at the same time.

Let's break down the motion of each rock separately:

Rock 1 (dropped rock):
- The initial velocity is zero (it's dropped from rest).
- The final velocity when it hits the ground is unknown.
- The acceleration due to gravity is -9.8 m/s² (negative because it acts in the downward direction).

Using the kinematic equation for vertical motion:
Final velocity (v) = Initial velocity (u) + Acceleration (a) x Time (t)

Since the rock is dropped from rest, the initial velocity (u) is zero. We can simplify the equation to:
v = a x t

For Rock 1, it will take some time (t1) to hit the ground.

Rock 2 (thrown rock):
- The initial velocity is v0 (thrown downward).
- The final velocity when it hits the ground is unknown.
- The acceleration due to gravity is -9.8 m/s² (again negative because it acts in the downward direction).

Using the same kinematic equation for vertical motion:
Final velocity (v) = Initial velocity (u) + Acceleration (a) x Time (t)

For rock 2, the initial velocity (u) is v0. We can write the equation as:
v = v0 + a x t

For Rock 2, it will take some time (t2) to hit the ground. The time difference between Rock 2 and Rock 1 is 1.2 seconds (t2 - t1 = 1.2 s).

Now we can equate the times and solve for v0:
t1 = t2 - 1.2 s

Plugging the equations for both rocks into the time equation:
a x t1 = v0 + a x t2

Since a (acceleration due to gravity) is the same for both rocks and can be factored out:
a x (t2 - 1.2 s) = v0 + a x t2

Simplifying the equation:
a x t2 - 1.2a = v0 + a x t2

Canceling out the a x t2 terms on both sides:
- 1.2a = v0

Finally, we can solve for v0 by substituting the acceleration due to gravity:
- 1.2 x (-9.8) = v0
v0 = 11.76 m/s

Therefore, the initial speed of the rock thrown downward is approximately 11.76 m/s.