Answer:
the body has descended a height (4a)
Explanation:
This exercise should use the acceleration given in the graph, but unfortunately the graph is not loaded, but we can build it, using the law of universal gravitation and the fact that you indicate that the movement is near the surface of the planet
F = m a
f
orce is gravitational force
G M m / r² = m a
a = G M / r²
where G is the universal constant of gravitation, M the mass planet and r the distance from the center of the planet of radius R to the body, if we measure the height of the body from the surface of the planet (y), we can write
r = R + y
for which
a = G M/R² (1+ y/R)⁻²
if we use y«R we can expand the function in series
a = (G M /R²) (1 -2 y/R - 2 (-2-1) /2! y² / R² +…)
as the height is small we can neglect the quadratic term and in many cases even the linear term, for this exercise we will remain only constant therefore the acceleration is constant
a = G M / R²
from this moment we can use the relations of motion with constant acceleration for the exercise
a) they ask us for the position for t = 2s
y = y₀ + v₀ t - a t²
as the body is released v₀ = 0
y-y₀ = - a t²
y-y₀ = - a 2²
y-y₀ = - a 4
therefore
therefore the body has descended a height (4a) where a is the acceleration of the planet's gravity
