1. The problem statement, all variables and given/known data 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? 2. Relevant equations One-Dimensional Kinematics 3. The attempt at a solution First I tried setting the distances of the balloon and the camera equal to each other, camera: 12t+.5*-9.81t^2 balloon: 2.3t+2.5 12t+.5*-9.81t^2=2.3t+2.5 then I rearranged the equation 12t-4.905t^2-2.3t=2.5 9.7t-4.905t^2-2.5=0 -4.905t^2+9.7t-2.5=0 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= -9.7+-6.71/9.81 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= 3.66-4.905*.093 x= 3.66 -.456 x= 3.2m 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.