A falling body has an initial velocity of 200 cm/sec. How far will it drop before it's velocity becomes 10,000 cm/sec.

Vo = 200 cm/s = 2 m/s.

V = 10,000 cm/s = 100 m/s.

V^2 = Vo^2 + 2g*d = !00^2.
2^2 + 19.6*d = 100^2.
Solve for d.

To find the distance a falling body will drop before its velocity becomes a certain value, we can use the equation of motion for uniformly accelerated motion.

The equation of motion is:
v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity
a = acceleration (in this case, it is acceleration due to gravity, which is approximately 980 cm/sec^2)
s = distance

In this case, the final velocity (v) is 10,000 cm/sec, the initial velocity (u) is 200 cm/sec, and the acceleration (a) is -980 cm/sec^2 (negative because it is directed downward).

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

Substituting the values, we get:

s = (10,000^2 - 200^2) / (2 * -980)
s = (100,000,000 - 40,000) / (-1,960)
s = 99,960,000 / -1,960
s ≈ -51,000 cm

The negative sign indicates that the distance is measured downward from the initial position. So, the body will drop approximately 51,000 cm before its velocity becomes 10,000 cm/sec.