Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

The values are


We must find the values of t such that the height of the camera is 0 (when it hits the ground)


Multiplying by 2

Clearing the coefficient of 

Plugging in the given values, we reach to a second-degree equation

The equation has two roots, but we only keep the positive root

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is



Answer:
if it were to be digital then you would be able to carry it around where ever and look at it to remind you of the memories, however it probably wouldnt last as long as the physical one. eg if you phone breaks there is a chance to restore the photo but you would had to have backed it up beforehand.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.