A particle is projected vertically upwards with a velocity of 34.3 m/s. For how long is the particle at a height of 49 m and above the point of projection?

Calculate the time to reach a height of 49 m:

S = V0 t - 1/2 g t^2

S = 34.3 t - 1/2 * 9.8 * t^2

4.9 t^2 - 34.3 t + 49 = 0

I get t = 2 and t = 5    as solutions of the quadratic.

We know that it takes v / a = 34.3 / 9.8 = 3.5 sec to reach maximum height.

So we have:

2 sec to reach 49 m

3.5 sec to reach max height

1.5 sec from 49 m to max height

1.5 sec from max height back down to 49 m

2 sec to fall from 49 m to ground

2 sec + 1.5 sec + 1.5 sec = 5 sec  to go from ground to max height and back down to 49 m

Option (a) is the correct answer.

Explanation:

Particles of a solid are held by strong intermolecular forces of attraction due to which they are not able to move from their initial place unless an external force or pressure is applied.

In liquids, molecules are less strongly held by each other as compared to solids and they are able to slide past each other. Therefore, liquids occupy volume of a container in which they are kept.

In gases, molecules are far apart from each other as they are held by weak Vander waal forces. Hence, particles of a gas have high kinetic energy and they are able to diffuse readily into the atmosphere or surroundings.

Thus, we can conclude that in the given situation a change from liquid to gas is taking place.