Answer:
5 m/s
Explanation:
Horizontal distance traveled, x = 2 m
vertical distance traveled, y = 4/5 m
Let the speed of cup as it leaves the counter is v and it takes time t to hit the ground.
Use second equation of motion in vertical direction
[tex]y = ut+\frac{1}{2}at^{2}[/tex]
Here acceleration in vertical direction is 9.8 m/s^2.
So,
[tex]\frac{4}{5} = 0+\frac{1}{2}\times9.8t^{2}[/tex]
t = 0.4 second
Now in horizontal direction the acceleration in zero.
Horizontal distance = horizontal velocity x time
x = v t
2 = v (0.4)
v = 5 m/s
Thus, the horizontal velocity of cup as it leaves the counter is 5 m/s.