Kinematics Problem on hot air balloon

[gadawg90]

A hot air balloon has just lifted off and is rising at the constant rate of 1.30 m/s. Suddenly, 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 9.40 m/s. If the passenger is 2.30 m above her friend when the camera is tossed, how much time does it take for the camera to reach her? How high is the passenger when the camera reaches her?

I am not completely sure how to even start this. I know you have to factor in that the air balloon is rising as she is throwing the camera in the air, but i do not know how to factor that inforation in. Help would be greatly appreciated. Thanks

 

[rl.bhat]

Hot air balloon is moving with constant acceleration. Camera is tossed with some initial velocity. When the passenger catches the camera, the distance traveled by her is x and the distance traveled by camera is x + 2.3 m. These distances were traveled in the same time interval. Using kinematic equations, find t and x.

 

[tomcue]

I can provide some guidance on how to approach this problem. Firstly, we need to understand the concept of kinematics, which is the study of motion without considering its causes. In this problem, we are dealing with two objects - the hot air balloon and the camera - that are moving in different directions and at different rates.

To solve this problem, we can use the equation of motion for an object in free fall, which is given by:

y = y0 + v0t + 1/2gt^2

Where:
y = final position
y0 = initial position
v0 = initial velocity
g = acceleration due to gravity (9.8 m/s^2)
t = time

In this case, we have two objects - the hot air balloon and the camera - moving in opposite directions. The hot air balloon is rising at a constant rate of 1.30 m/s, while the camera is being thrown upward with an initial velocity of 9.40 m/s. We can use this information to set up two separate equations for the vertical motion of each object.

For the hot air balloon, the equation will be:

y = y0 + v0t + 1/2gt^2

Where:
y0 = 2.30 m (initial position)
v0 = 1.30 m/s (initial velocity)
g = -9.8 m/s^2 (acceleration due to gravity, negative because the balloon is moving upward)
t = time (unknown)

For the camera, the equation will be:

y = y0 + v0t + 1/2gt^2

Where:
y0 = 0 m (initial position)
v0 = 9.40 m/s (initial velocity)
g = -9.8 m/s^2 (acceleration due to gravity, negative because the camera is moving upward)
t = time (unknown)

Now, we can use these equations to find the time it takes for the camera to reach the passenger in the hot air balloon. We can set the two equations equal to each other and solve for t:

2.30 + 1.30t + 1/2(-9.8)t^2 = 0 + 9.40t + 1/2(-9.8)t^2

Simplifying this equation, we get:

2.30 + 1.30t =