Answer:
The velocity of the camera is 33.11 m/s.
Explanation:
Given that,
Speed = 10.8 m/s
Altitude = 50 m
Suppose determine the velocity of the camera just before it hits the ground?
We need to calculate the velocity of the camera
Using equation of motion

Where, v = final velocity of camera
u = initial speed of camera
s = distance
Put the value into the formula



The direction will be downward so it is the negative velocity.
Hence, The velocity of the camera is 33.11 m/s.
I think the answer to your question is twenty-four percent.
Energy released during an earthquake travels in the form of waves<span> around the Earth. Two types of seismic </span>wave<span> exist, </span>P<span>- and </span>S-waves<span>. They are different in the way that they travel through the Earth. ... They are transverse </span>waves<span> which mean the </span>vibrations<span> are at right angles to the direction of travel.</span>
Explanation:
I believe <u>f</u><u>i</u><u>l</u><u>m</u><u> </u><u>i</u><u>s</u><u> </u>not part of a circuit
Answer:
Final velocity will be 314.6 m/sec
Distance traveled = 1314.24 m
Explanation:
We have given initial velocity u = 233 m/sec
Acceleration 
Time t = 4.8 sec
From first equation of motion
, here v is final velocity, u is initial velocity and t is time
So 
Now we have to find distance traveled
From second equation of motion

So distance traveled in given time will be 1314.24 m