Solving for Minimum Camera Speed to Reach Passenger

In summary: And yes, a=9.8 in this case.In summary, the problem involves a hot air balloon rising at a constant rate and a camera being thrown upwards from the ground towards the balloon. The goal is to find the minimum initial speed of the camera so that it reaches the passenger in the balloon. The position equations for the balloon and the camera are set up and it is determined that the two motions must intersect at the same time for the camera to reach the passenger. Using this information, the equation 2.7 + 1.87t = Vot - 4.9t^2 is derived. By solving for t and using the quadratic formula, the minimum initial speed of the camera, Vo, can be determined.
  • #1
BMWPower06
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0

Homework Statement


A hot air balloon has just lifted off and is rising at the constant rate of 1.87 m/s. One of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 11.6 m/s. If the passenger is 2.70m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger.


Homework Equations


Not really sure, but i was able to draw a FBD not exactly sure what else to do.
 
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  • #2
i don't understand, the initial speed of the camera is given?
 
  • #3
it gives an initial speed, but it wants to know the MINIMUM initial speed that the camera can be thrown.
 
  • #4
ok so do you know your position function?
 
  • #5
ice109 said:
ok so do you know your position function?

no, I am not sure i know what a position function is...
 
  • #6
[tex] s = s_o + V_o + \frac{1}{2}at^2[/tex]
 
  • #7
using the above position functions, write out the position equations for the balloon and for the camera.

Balloon: [tex]s_b = 2.7 + 1.87t[/tex]
Camera: [tex]s_c = V_ot - .5*g*t^2[/tex]

You want to find the minimum V_o such that [tex]s_b=s_c[/tex] has a solution for t > 0
 
  • #8
sam1 said:
using the above position functions, write out the position equations for the balloon and for the camera.

Balloon: [tex]s_b = 2.7 + 1.87t[/tex]
Camera: [tex]s_c = V_ot - .5*g*t^2[/tex]

You want to find the minimum V_o such that [tex]s_b=s_c[/tex] has a solution for t > 0

oooo you're going to get in trouble
 
  • #9
sam1 said:
using the above position functions, write out the position equations for the balloon and for the camera.

Balloon: [tex]s_b = 2.7 + 1.87t[/tex]
Camera: [tex]s_c = V_ot - .5*g*t^2[/tex]

You want to find the minimum V_o such that [tex]s_b=s_c[/tex] has a solution for t > 0
y does the first one (balloon) have 1.87t? shouldn't it just be 1.87, since a=0, so .5*0*t^2=0 right?

so for Sb=4.57 is that right?
then I am stuck on Sc, i have Sc= 0 + x + 4.9

what do i use for T?
ice109 said:
oooo you're going to get in trouble

what do u mean?
 
  • #10
do i plug in T= V - Vo/a

if so do i use 9.8 for a?

thanks
 
  • #11
He means that you aren't supposed to post solutions, and a moderator will delete the solution when he sees it.

The first balloon has 1.87t because you have a non-zero velocity term in there (which is 1.87).

How do you get that the balloon time 4.57 in the first place, and why are you not realizing that the two should be the same? Let me ask you this, if you want to spit on a car while standing on a bridge, do you find a spitting time versus a car time and then hope your spit suspends itself in time? No, you want the times to be the same so that the two motions intersect at the desired position, or target. Is this case really any different?
 
  • #12
ice109 said:
[tex] s = s_o + V_o + \frac{1}{2}at^2[/tex]

You missed a term. The equation should be..

[tex]s = s_o + V_o t + \frac{1}{2}at^2[/tex]
 
  • #13
Mindscrape said:
He means that you aren't supposed to post solutions, and a moderator will delete the solution when he sees it.

The first balloon has 1.87t because you have a non-zero velocity term in there (which is 1.87).

How do you get that the balloon time 4.57 in the first place, and why are you not realizing that the two should be the same? Let me ask you this, if you want to spit on a car while standing on a bridge, do you find a spitting time versus a car time and then hope your spit suspends itself in time? No, you want the times to be the same so that the two motions intersect at the desired position, or target. Is this case really any different?

ok i think i get it now.

so its:
2.7+1.87t=Vot-4.9t^2

right? but I am still a little lost as to what to plug in for t. If i use another equation i can do t= V-Vo/a
but what do i use for V?
And, does a=9.8?

Thanks
 
Last edited:
  • #14
^^so is this right?
 
  • #15
Yes, you got it.

Solve for t (you might want to consider the quadratic formula) and then you can figure out Vo.
 

Related to Solving for Minimum Camera Speed to Reach Passenger

What is the minimum camera speed required to reach a passenger?

The minimum camera speed required to reach a passenger depends on several factors such as the distance between the camera and the passenger, the speed of the passenger, and the angle at which the camera is moving. These factors can be used to calculate the minimum speed needed by using the laws of motion and kinematics.

How do you calculate the minimum camera speed to reach a passenger?

To calculate the minimum camera speed, you can use the equation: v = √(2ad/t), where v is the minimum speed, a is the acceleration, d is the distance, and t is the time. You will need to have accurate measurements of these variables in order to calculate the minimum camera speed accurately.

What is the role of acceleration in determining the minimum camera speed?

Acceleration plays a crucial role in determining the minimum camera speed needed to reach a passenger. The camera must accelerate at a certain rate in order to cover the distance between itself and the passenger in the allotted time. The higher the acceleration, the faster the camera can reach the passenger.

Can the angle of the camera affect the minimum speed required to reach a passenger?

Yes, the angle of the camera can affect the minimum speed needed to reach a passenger. If the camera is moving at an angle, it will have to cover a greater horizontal distance to reach the passenger compared to if it was moving directly towards the passenger. This means that a higher minimum speed will be required to reach the passenger at an angle.

What are some practical applications of calculating the minimum camera speed to reach a passenger?

Calculating the minimum camera speed to reach a passenger can be useful in various situations, such as designing and testing drone or helicopter cameras for search and rescue missions. It can also be applied in sports photography or videography, where the camera needs to follow a moving athlete. Additionally, it can be used in the development of autonomous vehicles or robots that need to navigate around obstacles and reach a target location efficiently.

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