A rock is dropped from rest, off of a 30m tall bridge. What is the rock's final velocity after it has fallen down 30m and hits the water below? Assume a negligible effect from air resistance.

-24.2487113m/s

Explanation:

vi=0, x=30, a=9.8, t=?, vf=?

x=vi*t+.5*a*t^2

30=.5*9.8*t^2

30=4.9*t^2

t=2.4743583s

vf=vi+a*t

vf=0+9.8*2.4743583

vf=24.2487113m/s

Then change it to negative because you are going down.

Below

Explanation:

Here are all the values that we are given :

Initial velocity Vi = 0 m/s

Vertical distance Dy = 30 m

All objects accelerate towards earth at a rate of 9.8 m/s^2, this is our acceleration

We can use this formula to find the rocks final velocity as it hits the water :

     finalvelocity^2 = initialvelocity^2 + 2(acceleration)(distance)

Plug in our values :

     Vf^2 = 0^2 + 2(9.8m/s^2)(30m)

     Vf^2 = 588

Square this number to get Vf

     Vf = 24.2487 m/s

Now we have to round this number to significant figures. In this case we have to round it to 1 sig fig as the the only number we are given, 30, has 1 sig fig :

     Vf = 20 m/s

Hope this helps! Best of luck <3

answer; ///(74) degrees is your answer..

nah.   i'm just sleepy.