Question

A rock is thrown horizontally of a bridge at 8.00 m/s. It hits the water's surface below 3.4 seconds later. How high is the bridge? How far from the bridge does it hit the water?

I was provided with the equations:
x= v * t and y= (Vi * t)1/2 * a * t^2

Answer 1

56.7m

Step-by-step explanation:

horizontal velocity does not matter in this situation because it stays constant (no unbalanced force acting upon it)

a=9.81m/s^2

t=3.4s

Vi=0m/s

d=Vi*t+1/2at^2

d=1/2at^2

d=1/2*9.81m/s^2*(3.4s)^2

d=4.905m/s^2*11.56s^2

d=56.7m