Kinematics Free Fall problems help

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SUMMARY

The discussion focuses on solving a kinematics problem involving a hot air balloon and a camera tossed upward. The balloon rises at a constant rate of 2.4 m/s, while the camera is thrown with an initial speed of 11.6 m/s from a height of 2.5 m below the balloon. The initial calculations incorrectly determined the time to reach maximum height instead of the time to intersect with the balloon's height. Correctly applying the kinematic equations will yield the height of the passenger when the camera reaches her.

PREREQUISITES
  • Understanding of kinematic equations for uniformly accelerated motion
  • Knowledge of initial velocity, acceleration due to gravity, and time
  • Ability to solve quadratic equations
  • Familiarity with the concept of relative motion in physics
NEXT STEPS
  • Review the kinematic equations: v = v0 + at and x = x0 + v0t + 1/2at^2
  • Learn how to calculate the time of flight for objects under gravity
  • Study the concept of relative motion to analyze problems involving multiple moving objects
  • Practice solving similar free fall problems to reinforce understanding
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Students studying physics, particularly those focusing on kinematics, as well as educators seeking to clarify concepts related to projectile motion and free fall.

P944
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Homework Statement


I am having a heck of a time trying to solve these problems. Please help if you can.

A hot air balloon has just lifted off and is rising at the constant rate of 2.4 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 11.6 m/s. If the passenger is 2.5 m above her friend when the camera is tossed, how high is she when the camera reaches her?

Homework Equations


v= v0 +a*t
x=x0 + vt - 1/2*a*t^2
x=x0 + v0*t +1/2at^2
v^2 =V0^2 +2a(x-x0)


The Attempt at a Solution


I tried to solve for time then use that variable to solve for distance
V=Vo + at = 0=11.6m/s +(-9.80m/s^2)*t
1.184s = time

x = x0 +vt - 1/2at^2
x = 2.5m + (1.184s)(11.6m/s) - 1/2(-9.80m/s^2)(1.184s)^2 = 23.10

I would greatly appreciate if someone could help me understand where i went wrong or understand how to approach this problem. thank you so much
 
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Hi P944! Welcome to PF! :smile:
P944 said:
I tried to solve for time then use that variable to solve for distance
V=Vo + at = 0=11.6m/s +(-9.80m/s^2)*t
1.184s = time

Nooo :redface: … that's the time it reaches maximum height (v = 0), you need the time it passes the balloon. :wink:
 

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