A rock is dropped of the side of a bridge and hits the water below 4s later.

Question

a. What was the rock’s velocity when it hit the water?

b. What was the rock’s average velocity as it fell?

c. What is the height of the bridge above the water?

Answer 1

c. use kinematics equation
y=vi*t+1/2at^2
vi= 0 (because rock was dropped)
a=-9.8m/s^2
t= 4s (time given)
plug in and you get your height
a. use kinematics equation
vf^2=vi^2+2ay
vi=0 (rock dropped)
a=-9.8
y= (whatever you got from part c)
plug in and solve
b. ?

Answer 2

To do this problem you just need to use a few simple equations:
V = V0 + at
X = X0 + V0*t + .5*a*(t^2)
and
V^2 = V0^2 + 2*a*(X - X0)
where V0 equals the initial velocity
X0 equals the initial height
t equals the time
a is just the acceleration (gravity in this case [9.81 m/s^2])
V is the final velocity
X is the final height

I think this should help you.