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Serhud [2]
3 years ago
7

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

Mathematics
1 answer:
NARA [144]3 years ago
4 0

Answer:

15.19 m/s

Step-by-step 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

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