A hot-air balloon has just lifted off and is rising at the constant rate of 2.3 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 12 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?
The Attempt at a Solution
First I tried setting the distances of the balloon and the camera equal to each other,
then I rearranged the equation
Next, I plugged the equation into the quadratic formula.
t= -9.7+-sqrt 9.7^2-4.905*-2.5/2*-4.905
t= -9.7+-sqrt 94.09+19.62*-2.5/-9.81
t= -9.7+-sqrt 94.09-49.05/-9.81
t= -9.7+-sqrt 45.04/9.81
t= .305, 1.67
next, I plugged the t value into the formula x= Xi + Vi*t + .5at^2
x= 0 + 12*.305 + .5*-9.81*.305^2
x= 3.66 -.456
I entered this into the website we turn our homework into, but it said the answer was incorrect, I'm not really sure what I did wrong, all my math seems correct.