Refer to the diagram shown below.
h = height of the girl above water when she lets go of the rope.
The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.
The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
= 8.3 m (nearest tenth)
Answer:
When the girl let go of the rope, she was about 8.3 m above the surface of the water.
Answer:
45.93°
Explanation:
The angle of incidence is given as 32.7°
The refractive index of the water that is 
Refractive index of the air that is
(because the refractive index of air is 1 )
We have to find the angle at which the light leave the water means angle of refraction
So according to snell's law 


r =45.93°
So the light leave the water at an angle of 45.93°
Answer:
B
Explanation:
Formula
Mechanical advantage = length of the ramp / height
Givens
Length of the ramp = 8
Mechanical advantage = 2
height = ?
Solution
2 = 8feet / height Multiply both sides by height
2* height = 8 feet * height / height Combine
2* height = 8 feet Divide by 2
2*height/2 = 8 feet/2
height = 4 feet.
Answer:
Of the three stars in the system, the dimmest - called Proxima Centauri - is actually the nearest star to the Sun. The two bright stars, called Alpha Centauri A and B form a close binary system; they are separated by only 23 times the Earth - Sun distance.