A ball dropped from a height of 125 m takes 5 s to hit the ground and is traveling at the rate of 50 m/s when it hits. After 3 seconds, the ball travels at a rate of ___? After 4 seconds?

41 30 m/s

To find the velocity of the ball after a specific time, we can use the kinematic equation:

v = u + at

where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.

Given:
u = 0 m/s (since the ball was dropped)
t = 3 s (for the first case)
t = 4 s (for the second case)

We need to find the acceleration (a) first. We can use another kinematic equation:

v^2 = u^2 + 2as

where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the distance traveled.

Given:
v = 50 m/s (final velocity when the ball hits the ground)
u = 0 m/s (initial velocity when the ball was dropped)
s = 125 m (height from which the ball was dropped)

By substituting these values into the equation, we can solve for a:

50^2 = 0^2 + 2a(125)
2500 = 250a
a = 10 m/s^2

Now, we can calculate the velocity after 3 seconds and 4 seconds.

For t = 3 s:
v = u + at
v = 0 + (10 m/s^2)(3 s)
v = 30 m/s

For t = 4 s:
v = u + at
v = 0 + (10 m/s^2)(4 s)
v = 40 m/s

Therefore, after 3 seconds, the ball will be traveling at a rate of 30 m/s, and after 4 seconds, it will be traveling at a rate of 40 m/s.