a marble 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?

Respuesta :

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

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