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A parcel is traveling on a horizontal conveyer belt moving at 1 meter/second. At the end of the conveyor belt, the parcel lands on a tray after 1 second. Calculate the horizontal and vertical distance of the tray from the end of the conveyor belt.

A)1.9 meters, 2.3 meters

B)1 meter, 4.9 meters

C)3.2 meters, 1.2 meters

D)3.3 meters, 4.3 meters

E)9.8 meters, 4.2 meters

Respuesta :

claytonbeast22
answer is B) 1 meter, 4.9 meters

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).