Agrasshopper jumps a horizontal distance of 1.50 m from rest, with an initial velocity at a 43.0° angle with respect to the horizontal.
(a) find the initial speed of the grasshopper.
m/s
(b) find the maximum height reached.
m

(a)v₀ =3.84 m/s

(b) h= 0.35 m

Explanation:

The grasshopper describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

g = 9.8 m/s²

v₀ , at an angle  α=43.0°  above the horizontal

xi=0

x=  1.50 m

y₀=0

x-y components of v₀

v₀x =  v₀cos43.0° = vx

v₀y =  v₀sin43.0°

Calculation of the time it takes for the grasshopperl to toch the ground

We replace data in the equation (1)

x = 0 + vx*t  =  (v₀cos43.0°)*t

t= x/(v₀cos43.0°)

t= 1,5/(v₀cos43.0°) Equation (3)

We replace t= 1,5/(v₀cos43°),y₀=0, y = 0,:when the grasshopper touches the ground  in the equation (2)

y=(v₀y)*t - (1/2)*g*t²

0=(v₀sin43°)*(1,5/(v₀cos43°) - (1/2)*(9.8)*1,5²/(v₀cos43°)²

0= 1,5*tan43° - 11.025/(v₀cos43°)²

11.025/(v₀cos43°)² = 1,5*tan43°

11.025 = (1,5*tan43°)*(v₀cos43°)²

(v₀cos43°)² = (11.025 )/ 1,5*(tan43°)

v₀² = (11.025 )/ (1,5*tan43°)(cos43°)²

v₀² = 14.73

v₀ = 3.84 m/s

Flight time

in the Equation (3):

t = 1,5/(v₀cos43.0°) = 1,5/(3.84*cos43°)

t = 0.534 s

(b) Maximum height reached (h)

The time to reach maximum height (h) is half the time of flight time :

th=0.534 s /2 = 0.267 s

We apply the  Equation (2)  

y= (v₀y)*t - (1/2)*g*t²

h= ( 3.84*sin 43°)*( 0.267) - (1/2)*(9.8)*( 0.267)²

h= 0.35 m

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.