Question

A watermelon 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?

Answer 1

Answer:

$Displacement=(17.890m)$ j

Explanation:

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

The initial speed is

$v_{0}=7.0\frac{m}{s}$

The final speed is

$v_{f}=20\frac{m}{s}$

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

$v_{f}^{2}=v_{0}^{2}+2gy$ (I)

Where $v_{f}$ is the final speed

Where $v_{0}$ is the initial speed

Where $g$ is the acceleration due to gravity

And where $y$ 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) :

$v_{f}^{2}=v_{0}^{2}+2gy$

$v_{f}^{2}-v_{0}^{2}=2gy$

$y=\frac{v_{f}^{2}-v_{0}^{2}}{2g}$

The value of g is $g=9.81\frac{m}{s^{2}}$ ⇒

$y=\frac{(20\frac{m}{s})^{2}-(7.0\frac{m}{s})^{2}}{2.(9.81)\frac{m}{s^{2}}}$

$y=17.890m$

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

$Displacement=(17.890m)$ j

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

Answer 2

Answer:

-18m

Explanation:

:)