A hot-air balloon has just lifted off and is rising at the constant rate of 2.2 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. If the passenger is 3.0 m 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?

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anonymous · Answer

Help please

anonymous · Answer

|dw:1346346688075:dw|Relevant equations assuming constant accelerations:
$$v = at+v _{0}$$$$x = \frac{ 1 }{ 2 }at ^{2}+v _{0}t+x _{0}$$
acceleration on the camera is due to gravity and will be downward
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Assumption is that the camera will just reach the balloon meaning that the final velocity of the camera will be zero. Now find an equation for how much time it would take for the camera to reach its peak. Use the two equations above, solve for V0 in the first and plug into 2nd and use the quadratic to find t. Then plug t back into equation 1 to find Vo

anonymous · Answer

I am still a little confused but I think I understand the basis of it! Thanks =)