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:

y = 17,89 m

Explanation:

Let us fixate the reference point in top of the building, from where the watermelon is thrown down. We will assume also that the positive axis of our system points up. We describe the watermelon’s motion with the equation:

 $v_y^2 =v_0^2 + 2ay$

Clearing the equation so we isolate y we have that:

 $y = (v_y^2 - v_0^2 )/2a$

Making a substitution with the values from the statement we have:

$y = ((20 m/s)^2 - (7 m/s)^2)/(2*9,81 m/s^2) = 17,89 m$]

So, this skyscraper is about 17,89 m tall; which is not very tall for a skyscraper but who am I to judge.  17,89 m is also the displacement of the watermelon from the point it was thrown down.

I hope everything was clear with my explanation. If I can help with anything else, just let me know. Have an awesome day :D

Answer 2

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.