Tourist takes a trip in hot air balloon . the hot air balloon is moving vertically upwards at a constant speed of 44m/s. at the height of 20m she drops the camera by accident over the side of the balloons basket. How do you calculate displacement covered by the camera after 2 seconds

68.4 m

Explanation:

We are given;

Height; y_o = 20 m

Speed; u = 44 m/s

Time; t = 2 s

To solve this, let's first of all find the position of the camera at a time of seconds after its release.

We will use one of Newton's equation of motion;

y(2) - y_o = ut - ½gt²

Negative sign is used on the right hand side because motion is against gravity.

Thus;

y(2) = y_o + ut - ½gt²

y(2) = 20 + (44 × 2) - ((1/2)(9.8)(2²))

y(2) = 88.4 m

Therefore displacement after 2 seconds will be;

y = 88.4 - 20

y = 68.4 m

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

newton's first law of motion