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
[tex]v^2=u^2+2gs[/tex]
Where, v = final velocity of camera
u = initial speed of camera
s = distance
Put the value into the formula
[tex]v^2=(-10.8)^2+2\times(-9.8)\times(-50)[/tex]
[tex]v=\sqrt{1096.64}[/tex]
[tex]v=33.11\ m/s[/tex]
The direction will be downward so it is the negative velocity.
Hence, The velocity of the camera is 33.11 m/s.
