You
drop
a
rock
off
a
bridge.
4.
(2
pts)
How
fast
is
it
going
after
2.0
s?
V = Vo + gt = 0 + 9.8*2 = 19.6 m/s.
To determine how fast the rock is going after 2.0 seconds, we need to use the equation for motion. The equation is:
v = u + at
Where:
v = final velocity (speed) of the rock
u = initial velocity (speed) of the rock (which in this case is 0 because the rock is dropped)
a = acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/s^2)
t = time (which is 2.0 seconds in this case)
By plugging in the values into the equation, we can solve for the final velocity (v). Let's do the math:
v = 0 + (9.8 m/s^2)(2.0 s)
v = 0 + 19.6 m/s
v = 19.6 m/s
After 2.0 seconds, the rock will be going at a speed of 19.6 m/s.