Let A be the second to last digit and let B be the last two digits of your 8-digit student ID. Example: for 20245347, A = 4 and B = 47.A ball rolls off a table. The table top is 1.2 m above the floor and the ball lands 3.6 m from the base of the table. Determine the speed of the ball at the time it rolled over the edge of the table? Calculate the answer in m/s and rounded to three significant figures.

7.35 m/s

Explanation:

Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.

y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.

So, substituting the values of the variables into the equation, we have

y - y' = ut - 1/2gt²

0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²

- 1.2 m = 0 - (4.9 m/s²)t²

- 1.2 m = - (4.9 m/s²)t²

t² = - 1.2 m/- (4.9 m/s²)

t² = 0.245 s²

t = √(0.245 s²)

t = 0.49 s

Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.

So, d = vt

v = d/t

= 3.6 m/0.49 s

= 7.35 m/s

Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.

It is shown thus V = √(u² + v²)

= √(0² + v²)

= √(0 + v²)

= √v²

= v

= 7.35 m/s

nah.   i'm just sleepy.