Awatermelon is thrown down from a skyscraper with a speed of 7.0\,\dfrac{\text m}{\text s}7.0 s m 7, point, 0, space, start fraction, m, divided by, s, end fraction. it lands with an impact velocity of 20\,\dfrac{\text m}{\text s}20 s m 20, space, start fraction, m, divided by, s, end fraction. we can ignore air resistance. what is the displacement of the watermelon?

j

Explanation:

First, let's write all the data for this exercise.

The initial speed is

The final speed is

If we ignore air resistance, we can use the following equation to calculate the distance travelled by the watermelon in this freefall motion :

(I)

Where is the final speed

Where is the initial speed

Where is the acceleration due to gravity

And where is the distance traveled by the object

The gravity acceleration has a positive sign if we consider as positive the sense of the free fall motion (downward positive sense).

If we replace all the data in the equation (I) :

The value of g is ⇒

Now, if we want to give a value to the displacement vector, we continue with our reference system (which we considered positive downward sense)

j

Where j is the unitary vector that gives to the displacement the vectorial character.

the pull of gravity on the rocket decreases