Answer:
5 m/s
Step-by-step 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
![y = ut+(1)/(2)at^(2)](https://img.qammunity.org/2020/formulas/physics/college/20gs654uuaf44u3zxo2xwmrmnw1ltyr2kv.png)
Here acceleration in vertical direction is 9.8 m/s^2.
So,
![(4)/(5) = 0+(1)/(2)*9.8t^(2)](https://img.qammunity.org/2020/formulas/physics/college/dljdaivcj0kwznbk66uj2jyh8i0c1rzzze.png)
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.