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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

Respuesta :

MathPhys

Answer:

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