Question

A hot air balloon has just lifted off and is rising at the constant rate of 1.80 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 11.5 m/s. If the passenger is 2.70 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?

Answer 1

For balloon,

X = 2.7 + 1.8t

For camera,

X = 11.5t - 4.9t^2

Comparing each other,

2.7 + 1.8t = 11.5t - 4.9t^2

4.9t^2 - 9.7t + 2.7 = 0

Time,t = 0.335 sec and t = 1.64 sec

B)

Distance, d = 2.7 + 1.8 x 0.335 = 3.30 m

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