Amarble rolls off a tabletop 1.0 m high and hits the floor at a point .54 m away from the table's edge in the horizontal direction. what is tge speed of the marble when it leaves the tables edge?

consider the motion of the marble in vertical direction or Y-direction :

Y = vertical displacement as the marble rolls from tabletop to floor = height of table = - 1 m                   (negative since the displacement is in down direction)

g = acceleration due to gravity = - 9.8 m/s²

t = time taken to travel from tabletop to bottom = ?

v₀ = initial velocity in vertical direction = 0 m/s

using the kinematics equation

Y = v₀ t + (0.5) a t²

- 1 = (0) t + (0.5) (- 9.8) t²

t = 0.45 sec

consider the motion of the marble in horizontal direction or X-direction :

X = horizontal displacement as the marble rolls from tabletop to floor = height of table = 0.54 m                  

a = acceleration in horizontal direction = 0 m/s²

t = time taken to travel from tabletop to bottom = 0.45 sec

v = initial velocity in horizontal direction = ?

using the kinematics equation

X = v t + (0.5) a t²

0.54 = v (0.45) + (0.5) (0.45) t²

v = 1.2 m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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