1. The problem statement, all variables and given/known data A softball is hit with an initial velocity of 29 m/s. at an angle of 60 degrees above the horizontal and impacts the top of the outfield fence 5s later. Assuming the initial height of the softball was .5 m above (level) ground, what are the balls horizontal and vertical displacements? 2. Relevant equations [itex] a_xt + V_i = V_f [/itex] [itex] \Delta x = (1/2) (V_f + V_i ) t [/itex] [itex] \Delta x = V_it + (1/2)a_xt^2[/itex] [itex] \Delta x = V_f - (1/2) a_xt^2 [/itex] 3. The attempt at a solution [itex] V_y = 29sin(60) = 25.11473671 m/s[itex] [itex] V_x = 29cos(60) = 14. 5 m/s [/itex] Horizontal displacement is easy. Using formula [itex] \Delta x = (1/2) (V_f + V_i ) t [/itex] with [itex] V_f = V_i [/itex] you get [itex] \Delta x = V_ft[/itex], so horizontal displacement = (14.5 m/s)(5s) = 72.5 m Now to find vertical displacement: I use formula [itex] a_xt + V_i = V_f [/itex] to find final velocity, plugging in I get: [itex] (-9.8 m/s^2) (5s) + 25.11473671 m/s = V_f [/itex] = [itex] -23.88526329 m/s = V_f [/itex] Now using [itex] \Delta y = (1/2)(V_f + V_i)(t) [/itex] I get: [itex] \Delta y = (1/2)(-23.88526329 m/s + 25.11473671 m/s) (5s) [/itex] = 3.07368355 m = y displacement But the softball starts off with in initial height of .5 m, so I subtract y displacement - .5, which = 2.57368355 m, but my book gives me answer: 2.95 m Can anyone tell me where I went wrong here?? I don't see my mistake..