Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 Time to height Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s Max height achieved is: H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m It falls that distance, minus Andrew's catch distance: h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m Time to descend is therefore: Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s Total time is rise plus fall therefore: Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)
Charge of balloon after 12,000 electrons have been removed from it = 12,000 × 1.6×10^-19 Coulomb = 1.92×10^-15 Coulomb = 1.92×10^-15/10^-6 = 1.92×10^-9 microCoulomb
We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :
