I'm sorry but I don't see anything that says part B
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When light crosses the boundary between layers with different densities, the light is refracted. (A).
Answer:t=0.3253 s
Explanation:
Given
speed of balloon is 
speed of camera 
Initial separation between camera and balloon is 
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is


and distance travel by camera to reach balloon is


Now






There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is 
velocity is given by



and position of camera is same as of balloon so
Position is 

Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
Explanation:
Initial velocity u = V₀ in upward direction so it will be negative
u = - V₀
Displacement s = H . It is downwards so it will be positive
Acceleration = g ( positive as it is also downwards )
Using the formula
v² = u² + 2 g s
v² = (- V₀ )² + 2 g H
= V₀² + 2 g H .
v = √ ( V₀² + 2 g H )