A rock is dropped from a high bridge. At the end of 3sec of free fall. The speed of rock is, in centimeters per second

if a = -g

then
v = Vi - g t
h = Hi + Vi t - (1/2) g t^2
so
v = 0 - 9.81 * 3
so speed which is |v| is about 30 m/s

To calculate the speed of the rock at the end of 3 seconds of free fall, we can use the formula for free fall:

v = gt

Where v is the velocity or speed, g is the acceleration due to gravity, and t is the time. The acceleration due to gravity is approximately 980 cm/s^2.

Plugging in the values:

v = (980 cm/s^2) * (3 s)
v = 2940 cm/s

Therefore, the speed of the rock at the end of 3 seconds of free fall is 2940 centimeters per second.

To calculate the speed of the rock after 3 seconds of free fall, we need to consider the acceleration due to gravity. On Earth, this value is approximately 9.8 meters per second squared (9.8 m/s^2).

To convert this value to centimeters per second, we can multiply it by 100, which gives us 980 centimeters per second squared (980 cm/s^2).

After 3 seconds, the velocity of the rock can be calculated using the formula: velocity = initial velocity + (acceleration × time).

However, since the rock is initially at rest (dropped from rest), the initial velocity is 0 cm/s. Therefore, the formula simplifies to: velocity = acceleration × time.

Plugging in the values, we get: velocity = 980 cm/s^2 × 3 s.

Multiplying these values together gives us a final answer of 2940 cm/s. Hence, the speed of the rock at the end of 3 seconds of free fall is 2940 centimeters per second.