An object is dropped near the surface of a planet such that the gravitational field at the object’s location is 8 nkg. if the object is thrown upward at a speed of 20 ms, what is the position of the object in relation to the position in which the object was released and thrown upward after 3s ?
a- 96 m below the release position
b- 36 m below the release position
c-15 m above the release position
d- 24 m above the release position

D- 24 m above the release position

Explanation:

Given:

v₀ = 20 m/s

a = -8 m/s²

t = 3 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20 m/s) (3 s) + ½ (-8 m/s²) (3 s)²

Δx = 24 m

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therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

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