Amarble rolls off a tabletop 1.0 m high and hits the floor at a point 1.2 m away from the table’s edge in the horizontal direction. what is the speed of the marble when it leaves the table’s edge?
a. 1.2 m/s
b. 9.8 m/s
c. 2.7 m/s
d. 3.7 m/s

C. 2.7 m/s

Explanation:

Consider the motion of the marble along the vertical direction

y = vertical displacement = height of tabletop = 1.0 m

a = acceleration = 9.8 m/s²

t = time taken to hit the floor = ?

v₀ = initial speed of the marble along vertical direction = 0 m/s

Using the equation

y = v₀ t + (0.5) a t²

1.0 = (0) t + (0.5) (9.8) t²

t = 0.45 s

Consider the motion of the marble along the horizontal direction :

v = speed with which the marble leaves the table

x = horizontal displacement = 1.2 m

t = 0.45 s

horizontal displacement  is given as

x = v t

1.2 = v (0.45)

v = 2.7 m/s

see earlier answer