At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building that is 300 m tall. how much earlier does the thrown rock strike the ground? neglect air resistance. show all work and formulas used

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

So now we have an equation and unkown value.

for the thrown rock

for the dropped rock

solving both equation with the quadratic formula:

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

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.

i would say