Physics Motion?

Question

Hello all again, well i am having a problem with motion. Here is the problem. You want to know how high a bridge is, so you go out to the middle of the bridge with a stone and a stopwatch. You drop the STONE over the side. It hits the water 4.6 seconds later. How high is the bridge?

OK well at first i was just thinking of multiplying 9.8m/s(2) and 4.6(s) but i don't really think that is right and this is where i am kind of stuck now. Can anyone help?

Answer 1

use basic kinematics equation:

(1) v = v0 + a*t
(2) v^2 = v0^2 - 2*a*(x-x0)

v0 = 0, a = 9.8 m/s^2, t = 4.6s
solve for v in first equation

x0 = 0 in second equation, solve for x, should be negative

Answer 2

You can start by finding the final velocity of the rock

A = V/t ; V = 9.8 * 4.6 = 45.08

This is final velocity. To find the distance the rock traveled you must find average velocity and multiply it by time again.

V(ave) = (Vf - Vi)/2 = (45.08 - 0)/2 = 22.54

D = V(ave) * t = 22.54 * 4.6 = 103.7m

Now you can make one simple formula that relates acceleration and time to distance by making some algebraic substitutions.

Basically:
D = (1/2) * at^2 (if you start with Vi = 0)
D = (1/2)*at^2 + Vi*t (if Vi is not = 0)