What Is the Maximum Height Reached by a Falling Camera from a Blimp?

  • Thread starter Thread starter joehoy41
  • Start date Start date
  • Tags Tags
    Camera Falling
AI Thread Summary
A camera falling from a blimp at 349 meters and rising at 17.2 m/s has an initial upward velocity of 17.2 m/s, which becomes zero upon release. The maximum height reached by the camera is 349 meters, as it starts falling from that point. The discussion includes various equations of motion to calculate the camera's speed upon hitting the ground. The acceleration due to gravity is 9.8 m/s², and participants clarify the initial conditions and relevant equations. Understanding these parameters is crucial for solving the problem accurately.
joehoy41
Messages
8
Reaction score
0
A camera falling from a blimb... pleasezzz help

(part 1 of 2)
A camera falls from a blimp that is 349 m above the ground and rising at a speed of 17.2 m/s.
Find the maximum height reached by the camera with respect to the ground. The acceleration of gravity is 9.8 m/s^2 Answer in units of m.

(part 2 of 2)
Find the speed at which it hits the ground. Answer in units of cm/s.
 
Physics news on Phys.org


So which equations do you know that relate to this?
 


there are a lot.. but i only want to know what does this mean "A camera falls from a blimp that is 349 m above the ground and rising at a speed of 17.2 m/s." can u give me a simple explanation of what they are saying?
 


Can someone pleasez help me
 


Part A of the question is improperly formed. Or it is trivial. Or you have written it out wrong.

As written, the maximum height the camera reaches is the point where it is released - 349m. Period.

Please re-examine the question you were given.
 


So, when the camera falls out of the rising blimp, its upward velocity immediately goes to zero?
 


i am sure that is the question...
 


SteamKing said:
So, when the camera falls out of the rising blimp, its upward velocity immediately goes to zero?
Ah, of course. My mistake. Good catch!

So the camera's params are:
Initial distance: 0m
Final distance: 349m
Initial velocity: -17.2m/s
Acceleration: +9.8m/s^2

joehoy, do you know your SAVTU equations?
 


Vav = (Df - Di)/t
Aav = (Vf - Vi)/t
Vf = Vi + gt
Vf = -sqrt(Vi^2 - 2gd)
d= 0.5(Vi + Vf)t
d = Vit + 0.5gt^2
avgt = (Vf - Vi)/g
t = sqrt((2d)/g)

g = 9.81 m/s^2
 
  • #10


joehoy41 said:
Vav = (Df - Di)/t
Aav = (Vf - Vi)/t
Vf = Vi + gt
Vf = -sqrt(Vi^2 - 2gd)
d= 0.5(Vi + Vf)t
d = Vit + 0.5gt^2
avgt = (Vf - Vi)/g
t = sqrt((2d)/g)

g = 9.81 m/s^2

OK. So you've got initial velocity and acceleration.

The first part of the question is only asking for the short arc of the camera as it rises to its peak.
So final velocity of that short arc is going to be what?

Which equation will give you distance, given initial v, final v and a?
 
Back
Top