Explanation :
It is given that,
Horizontal velocity of the conveyer belt, v = 1 m/s
At the end of the conveyor belt, the parcel lands on a tray after 1 second. We need to find the horizontal and vertical distance of the tray from the end of the conveyor belt.
For horizontal distance, d = v × t
d = 1 meters
For vertical distance, use the second equation of motion as :
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Initial vertical velocity, u = 0
[tex]s=0+\dfrac{1}{2}9.8\ m/s^2\times (1\ s)^2[/tex]
s = 4.9 meters
So, the horizontal and vertical distance of the tray from the end of the conveyor belt are 1 meters and 4.9 meters respectively. Hence, the correct option is (b).
