Final answer:
To calculate the velocity of a rock dropped from a bridge after 2.5 seconds, the equation v = u + at is used, where u is 0 m/s (since it was dropped) and a is 9.8 m/s² due to gravity. The calculation gives a final velocity of 24.5 m/s directed downwards.
Step-by-step explanation:
The subject of this question is Physics, specifically kinematics, which is focused on the motion of objects. To find the rock's velocity after it has been dropped from a bridge for 2.5 seconds, we can use the equation of motion for an object under constant acceleration due to gravity (neglecting air resistance). The equation is:
v = u + at
where:
- v is the final velocity,
- u is the initial velocity (for an object that is dropped, this is 0 m/s),
- a is the acceleration (which for gravity is approximately 9.8 m/s² downward), and
- t is the time the object has been falling.
Plugging in the values for a rock dropped from a bridge:
v = 0 m/s + (9.8 m/s²)(2.5 s)
v = 24.5 m/s
Therefore, the velocity of the rock after 2.5 seconds is 24.5 m/s directed downwards towards the water.