Now let's examine the case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in free fall and we can use the following equations to analyze its motion. Vy y = Voy - gt =30+ voyt-gt = vy - 29(y-yo) Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand; (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the effects of the air. (Figure 2) shows graphs of position and velocity as functions of time for this problem. Note that the graph of vy, versus t has a constant negative slope. Thus, the acceleration is negative on the way up, at the highest point, and on the way down. Part A Practice Problem: - If the building is 20.0 m high, how long does it take for the ball to hit the ground? Express your answer in seconds. VO ? Figure The ball actually moves straight up and then straight down, we show a U-shaped path for clarity. = 0 = ? f = 1.00 s t = ? y = ? 1 of 2 > Submit Request Answer Part B Practice Problem: S 1=7y = 5.00 m What velocity is the ball traveling at when it hits the ground? Express your answer in meters per second. f = 0, voy = 15.0 m/s y = 0 ay = 8 =-9.80 m/s VO t = 4.00 s U = ? y = ? Submit Request Answer Provide Feedback ? m/s Next