Answer:
15.19 m/s
Step-by-step explanation:
[tex]v_f =v_i + at[/tex], where [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] 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:
[tex]v_f = 0 + at[/tex]
We can plug in 9.8 m/s^2 into a and 1.55 s into t:
[tex] v_f = 9.8(1.55) [/tex]
[tex]v_f [/tex]= 15.19 m/s
