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.

convex lenses are thicker at the middle. rays of light that pass through the lens are brought closer together (they converge). a convex lens is a converging lens. when parallel rays of light pass through a convex lens the refracted rays converge at one point called the principal focus.