#$&* A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below. • For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction? #$&* Vertical direction: ⁃ v0 = 20 cm/s ⁃ `ds = 120 cm ⁃ a = 9.8 m/s^2 ⁃ vf^2 = v0^2 + 2 (980cm/s^2) ( 120cm) ⁃ = 400cm^2/s^2 + 235200cm^2/s^2 ⁃ = 235600 cm^2/s^2 vf = 485.4 cm/s vAve = (485.4 cm/s - 20 cm/s) / 2 = 232.7 cm/s `dt = (485.4 cm/s - 20 cm/s) / 980 cm/s^2 = 0.47 s • What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction? #$&* vf = 485.4cm/s, ⁃ `ds = 232.7 cm/s * 0.47 s = 109.4 cm in the vertical direction `dy ⁃ `dv = 485.4 cm/s - 20 cm/s = 465.4 m/s ⁃ vAve = 465.4 cm/s / 2 = 232.7 cm/s • What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval? #$&* Horizontal ⁃ a = 0 cm/s^2 (gravity has taken over) ⁃ v0 = 80 cm/s • What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval? #$&* `ds = 80 cm/s * 0.49 s = 39.2 cm ⁃ vf = 0 cm/s because theres no acceleration in the horizontal axis anymore ⁃ vAve = 80 cm/s (due to zero acceleration) • After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated? #$&* Yes because the ball was in free fall at an acceleration equal to the force of gravity in which the vf was greater than the v0 over a positive change in position. So the ball accelerated at a constant rate. • Why does this analysis stop at the instant of impact with the floor? #$&* At the point of impact, acceleration has stopped, along with velocity so there is no motion.