I suppose I am having trouble with this problem because there are no numbers given. Here is the problem: A ball is thrown down vertically with an initial speed of Vo from a height of h. a) What is its speed just before it strikes the ground? Speed is just the magnitude of instantaneous velocity, right? So I just used the equations for velocity: Vf =? V^2 = Vo^2 + 2a(Y-Yo) Since I'm dealing with free fall accel, a = -9.8 m/s^2. In addition, Y-Yo = h. So, V^2 = Vo^2 + 2(-9.8)h --> V^2 = Vo^2 -19.6h Leaving me with: V = (Vo^2 -19.6h)^(1/2) b) How long does the ball take to reach the ground? I used: Y-Yo = Vot + (1/2)at^2 and using the quadratic equation, I got: ** i used the "+" version bec t cannot be negative t = [-Vo + (Vo^2 - 19.6)^(1/2)]/-9.8 c) and d) are the same as a) and b) except this time the ball is thrown upward. I also have to figure out if the values for c and d are less than, greater than, or the same as the values in a and b. <b>c)</b> I used: Y - Yo = (1/2)(Vo + V)t h = (1/2)(Vo + V)t V = (2h/t) - Vo I think the values will be the same. d) For this part, I used the V found in c) V = (X1 - X)/t1 - t --> V = h/t --> t = h/V this is the time just to fall from the maximum point the ball reaches, so the total time to fall to the ground would be: 2t = 2(h/V) Because the time is doubled, I think this value will be greater than the value in b) Were my assumptions correct?