Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring compressed by a distance d = 0.21 m at an angle of θ = 32° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

Since we only care about the velocity , we can keep only the second and third parts of the equation and solve:

So, the speed of the ball just after the launch is 17.3m/s.

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 think its d if wrong im sorry