A ball is dropped from a bridge 122.5 m above a river. After the ball has been falling for 2s, a second ball is thrown straight down after it. What must the initial velocity of the second ball be so that both hit the water at the same time?

Let's analyze the motion of each ball separately.

Ball 1 (dropped):

• Initial velocity (v1_i) = 0 m/s
• Acceleration (a) = 9.8 m/s² (due to gravity)
• Distance (d) = 122.5 m

We can use the following kinematic equation to find the time it takes for ball 1 to hit the water:

d = v1_it + (1/2)at²
122.5 = 0t + (1/2)9.8t²
122.5 = 4.9*t²
t² = 122.5/4.9
t = √(122.5/4.9) ≈ 5 s

So, ball 1 takes approximately 5 seconds to hit the water.

Ball 2 (thrown):

• Time to hit the water (t2) = 5 s - 2 s = 3 s (because it's thrown 2 seconds later and hits at the same time)
• Distance (d) = 122.5 m
• Acceleration (a) = 9.8 m/s²
• We need to find the initial velocity (v2_i)

Using the same kinematic equation:

d = v2_it2 + (1/2)at2²
122.5 = v2_i3 + (1/2)9.83²
122.5 = 3v2_i + 44.1
3v2_i = 122.5 - 44.1
3*v2_i = 78.4
v2_i = 78.4/3 ≈ 26.13 m/s

Therefore, the initial velocity of the second ball must be approximately 26.1 m/s for both balls to hit the water at the same time.