If a rock were dropped from a height of 0.60 m above the comet's surface, how long would it take to hit the surface?

This question is incomplete, the complete question is;

In 2014, the Rosetta space probe reached the comet Churyumov-Gerasimenko.

Although the comet's core is actually far from spherical, in this problem we'll model it as a sphere with a mass of 1.0 × 10¹³ kg and a radius of 1.6 km .

If a rock were dropped from a height of 0.60 m above the comet's surface, how long would it take to hit the surface?

its will take approximately 67.845 seconds for the rock to hit the surface.

Explanation:

Given the data in the question;

mass of comet M = 1.0 × 10¹³ kg

radius r = 1.6 km = ( 1.6 × 1000 )m = 1600 m

height from which rock is dropped from s = 0.6 m

Now, acceleration due to gravity of the comet near the surface will be;

g = ( G × M ) / r²

we know that Newton's gravitational constant G = 6.67408 × 10⁻¹¹ m³.kg⁻¹s⁻²

so we substitute

g = [ (6.67408 × 10⁻¹¹) × (1.0 × 10¹³) ] / [1600]²

g = 667.408 / 2560000

g = 0.0002607 m/s²

from the equation of motion

s = ut + at²

initial velocity u will be 0

we substitute

0.6  = (0 × t) + [ × 0.0002607 × t² ]

0.6 = 0.00013035 × t²

t² = 0.6 / 0.00013035

t = √( 0.6 / 0.00013035 )

t = 67.845 seconds

Therefore, its will take approximately 67.845 seconds for the rock to hit the surface.

the answer is b i think