A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m? b. How long will it be in the air?

A. 29.7 m/s

B. 6.06 s

Explanation:

From the question given above, the following data were obtained:

Maximum height (h) = 45 m

A. Determination of the initial velocity (u)

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Initial velocity (u) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 45)

0 = u² – 882

Collect like terms

0 + 882 = u²

882 = u²

Take the square root of both side

u = √882

u = 29.7 m/s

Therefore, the ball must be thrown with a speed of 29.7 m/s.

B. Determination of the time spent by the ball in the air.

We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) to reach the maximum height =?

h = ½gt²

45 = ½ × 9. 8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:

Time (t) to reach the maximum height = 3.03 s

Time (T) spent by the ball in the air =?

T = 2t

T = 2 × 3.03

T = 6.06 s

Therefore, the ball spent 6.06 s in the air.

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.