Suppose our experimenter repeats his experiment on a planet more massive than earth, where the acceleration due to gravity is g=30 m/s2. when he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on earth? ignore friction and air resistance.
a. it will take more time to return to the point from which it was released.
b. it will smash his face. it will take less time to return to the point from which it was released.
c. it will stop well short of his face.
d. its mass will be greater.

B It will takes less time

Explanation:

Let's analyze this problem as a free fall with a greater acceleration of gravity, let's calculate the time it takes to reach the ground

     Y = Vo + + ½ at2

In this case Vo is zero and a = -30 m / s2

     Y = - ½ to t²

     t= √2y / a

     t =√ 2y / 30

Earth's time is t2 = √2y / 9.8

      t < t2

We can see that the time it takes for the ball to reach the floor is reduced, the rise time is the same as the drop.  The total time would be 2t on the massive planet and on Earth 2t2

So by comparing the results with those obtained on Earth the ball takes less time to return to the point of departure

0.55x745.7j and a calculator

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.