posted by nick .
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 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 on its way up?
Passenger height = 2.5m + vt
= 2.5m + (2.2m/s)t
Camera height = (11.6m/s)t + (1/2)(-g)t^2
The camera reaches the passenger when
Passenger height = Camera height
solve t for equal height, then plug t into either equation for the height.
You will find 2 answers for equal height. One is the time going up, and the second is if the time the camera would be at the passenger's height coming down (if the passenger missed the camera on its upward flight).