Question

A ball is dropped off the side of a bridge.
After falling 1.55 s, what is its velocity?

Answer 1

15.19 m/s

Explanation:

`v_f =v_i + at`, where
`v_f` is the final velocity,
`v_i` is the initial velocity, a is the constant gravitational acceleration 9.8 m/s^2, and t is the time (seconds).

Since the ball is dropped off the bridge, its initial velocity is 0:

`v_f = 0 + at`

We can plug in 9.8 m/s^2 into a and 1.55 s into t:

`v_f = 9.8(1.55)`

`v_f`= 15.19 m/s