The time of flight ends once the height of the ball hits the ground. First we look at the ball's upward trajectory. Its initial vertical velocity is (120 ft/s)(sin 30) = 60 ft/s. Since gravity is -32 ft/s^2, this means that the ball will fly upward for 60/32 = 1.875 s. By then, it will be at a height of: h = h0 + vi*t - 0.5gt^2 = 3 ft + (60 ft/s)(1.875 s) - 0.5(32 ft/s^2)(1.875)^2 = 59.25 ft So the time of its downward flight is solved by: 59.25 = 0.5(32)t^2 t = 1.924 s Total time of flight = 1.875 + 1.924 = 3.799 s The ball's horizontal velocity is constant at (120 ft/s)(cos 30) = 103.923 ft/s. Multiplying this by the time of flight of 3.799 s gives a total distance of 395 ft.
The fraction 6/11 is equivalent to the fraction of 18/33. They both have the same value. Simply multiply both the numerator and denominator by 3 to the fraction of 6/11 to get the new equivalent one.
