Posted by **Jackie** on Tuesday, April 6, 2010 at 12:19am.

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

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?

- Physics -
**echem**, Tuesday, April 6, 2010 at 12:26am
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. ?

- Physics -
**Fred**, Tuesday, April 6, 2010 at 12:31am
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.

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