course Phy 231 First we need to look at what the projectile is doing in the vertical direction.It's initial vertical velocity is 0, it will have a downward acceleration of 980m/s. We know that it falls 149cm so we will take the downward direction to be positive. We can find its final velocity in the vertical direction by using the formula: Vf^2 = Vo^2 + 2a`ds Vf^2 = 0 + 2(980)(149) Vf = 540cm/s We can now find the time that it was in the air with the equation: Vf = Vo + a`dt 540 = 0 + 980`dt `dt = 0.55s We now look at what it's doing in the horizontal direction. The velocity in the horizontal direction may change, but it will be a negligible amount so we consider it to be the same. We know initial and final velocities and the time interval so to find the distance we use the equation: `ds = ( (Vo + Vf) /2 ) `dt `ds = ( (37 + 37) /2 ) 0.55 `ds = 20.35 The projectile traveled about 20.35cm in the horizontal direction while it was in the air."