1. The problem statement, all variables and given/known data There are four different components to this problem. Each component is short -- so don't worry. a. Suppose you throw a 0.4 kg ball straight upwards from a height of 1.5 m, at a speed of 4.0 m/sec. How much work have you done on the ball by accelerating it from rest to 4.0 m/s. Is this work positive or negative? b. How high do you expect the ball to get above the ground, if there is no air resistance? c. At what speed do you expect the ball in #5 to be going when it comes back to the height at which you threw it (1.5 m)? d. When the ball comes back down to 1.5 m, you find that it is moving at 3.6 m/sec. You correctly surmise that you can't neglect air resistance here. How much work has "air resistance" done on the ball? 2. Relevant equations x = x0 + voxt + (1/2)(axt2 v2x = v20x + 2ax (x - x0) vx = vox + axt WTOT = K2 - K1 3. The attempt at a solution a. WTOT = K2 - K1 (1/2)(m)(v2) = K (1/2)(0.4)(4.0) = 3.2 J = K2. K1 = 0 (initial velocity is zero) WTOT = 3.2 J b. vx = vox + axt 0 = (4.0) + (-9.8)(t) -4.0 = (-9.8)(t) -4.0 / -9.8 = t = .408 My thinking here is that when the ball reaches its maximum height, the velocity is zero. x = x0 + voxt + (1/2)(axt2 x = 1.5 + (4.0)(.408) + (1/2)(-9.8)(.408)2 x = 2.3163 c. I'm pretty sure this is just plain wrong. Not sure why, but I have that feeling. x = x0 + voxt + (1/2)(axt2 1.5 = 2.3163 + 0 + (1/2)(9.8)(t2) t = .408 (time to go up that distance is the same as the time to go down??) vx = vox + axt vx = 0 + (9.8)(.408)2 = 1.63 m/s d. I know I've screwed up already before. If the ball is moving at 3.6 m/s when it is coming back down WITH air resistance, thats already a bad sign.