How much time will it take for the cannonball to reach the ground and at what height will the ball be after each second of travel if its launched horizontally from the top of a 78.4-meter high cliff?

4 seconds-it will still take 4 seconds to fall 78.4 meters use the equation y = 0.5 g t2 and substitute -9.8 m/s/s for g. the vertical displacement must then be subtracted from the initial height of 78. 4 m. at t = 1 s, y = 4.9 m (down) so height is 73.5 m (78.4 m - 4.9 m ) at t = 2 s, y = 19.6 m (down) so height is 58.8 m (78.4 m - 19.6 m ) at t = 3 s, y = 44.1 m (down) so height is 34.3 m (78.4 m - 45 m) at t = 4 s, y = 78.4 m (down) so height is 0 m (78.4 m - 78.4 m) note: the cannon balls horizontal speed does not affect the time to fall a vertical distance of 78.4 m.