Cleo and Clare are looking from their balcony to a swimming pool below that is located 15 m horizontally from the bottom of their building. They estimate the balcony is 45 m high and wonder how fast they would have to jump horizontally to succeed in reaching the pool. What calculations would you show to help them determine the answer? Evaluate the practicality of their being able to succeed at jumping into the pool.

- anonymous

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- anonymous

@iGreen

- anonymous

@Michele_Laino

- anonymous

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## More answers

- anonymous

In three scenarios a truck is towing a boat on a trailer. Using 3 – 4 complete sentences indicate which scenario(s) (if any) is the force of the truck on the trailer greater than the force of the trailer on the truck:
Scenario A: masstruck = 1000 kg, massboat and trailer = 500 kg; velocity = 20 m/s
Scenario B: masstruck = 800 kg, massboat and trailer = 800 kg; velocity = 20 m/s
Scenario C: masstruck = 500 kg, massboat and trailer = 1000 kg; velocity = 15 m/s

- Michele_Laino

hello!

- Michele_Laino

have you got any ideas?

- anonymous

Not at all, i got some some earlier but theres a few that I'm lost on.

- Michele_Laino

I think that in all of your three cases, both forces have the same magnitude, since we have to apply the third law of Newton

- Michele_Laino

furthermore, since the velocity is constant, then we can consider the system truck+boat, an inertial system

- Michele_Laino

and, as you know, the laws of Physics are the same into all inertial systems

- anonymous

so none of them are greater than the other? correct?

- Michele_Laino

yes! correct, since the velocity of the system truck+boat, is constant in all of your three cases

- anonymous

how could i try and explain that in a few sentences?

- Michele_Laino

here is my sentence:
"since the velocity of the system truck+boat" is constant, as vector, then the same system "truck+boat" is an inertial system. Now within an inertial system, all law of Physics, have the same shape, as formula, so we can apply the third law of Newton, and we can conclude that the force applied by truck on boat have the same direction and magnitude of the force applied by the boat on truck. Furthermore those forces have opposite orientation each other"

- anonymous

what about the first question?

- Michele_Laino

we have to consider these two equations:
\[\left\{ \begin{gathered}
h = \frac{1}{2}g{t^2} \hfill \\
\hfill \\
d = vt \hfill \\
\end{gathered} \right.\]
where d=15 meters and h= 45 meters

- Michele_Laino

now, I solve the second equation for t, and I get:
\[t = \frac{d}{v}\]

- Michele_Laino

Now I substitute into the first equation, and I get:
\[45 = \frac{1}{2}g{\left( {\frac{{15}}{v}} \right)^2}\]
please solve that equation for velocity v, here g=9.81 m/sec^2

- anonymous

i dont understand you math I'm confused

- anonymous

@Michele_Laino

- Michele_Laino

the subsequent drawing represents the situation of your exercise:
|dw:1432832271358:dw|

- Michele_Laino

this time:
\[t = \sqrt {\frac{{2 \times 45}}{{9.81}}} \]
is the time needed to Cleo and Clare to reach the base of their hotel. During that time they have to travel, horizonatally by a distance L=15 meters, in order to reach the swimming pool, and the time needed to travel that distance L, is:
\[{t_1} = \frac{{15}}{v}\]
where v is the horizonatl velocity of Cleo and Clear.
Now we have to equate those two times, so we can write:
\[\sqrt {\frac{{2 \times 45}}{{9.81}}} = \frac{{15}}{v}\]
|dw:1432832762849:dw|

- Michele_Laino

so, what is v?

- anonymous

i dont understand the equation like how its written so i dont know

- Michele_Laino

do you know the law of falling bodies?

- anonymous

no

- anonymous

would the answer be 3.03 seconds ?

- anonymous

A bug and the windshield of a fast-moving car collide. Indicate whether each of the following statements is True or False (1 point each)
• The forces of impact on the bug and on the car are the same magnitude.
• The impulses on the bug and non the car are the same magnitude
• The changes in speed of the bug and of the car are the same magnitude.
• The changes in momentum of the bug and of the car are the same magnitude.
• The accelerations of the bug and the car are the same magnitude.

- anonymous

@Michele_Laino

- Michele_Laino

please wait a moment

- anonymous

okay :)

- Michele_Laino

the answer to the first question is:
\[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15\]

- anonymous

45=4.9t^2
9.184=t^2
t=3.03sec
That is what i got for the first

- Michele_Laino

yes I know, nevertheless, you have to compute the velocity v, since your problem asks for a velocity

- anonymous

so it that right?
and the 2nd one i got true, true, false, true, false?

- Michele_Laino

we have:
\[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15 = 4.95\;m/\sec \]

- anonymous

could you right the equation using the thing on here? thats the part that is getting me a little mixed up

- Michele_Laino

I'm pondering....

- Michele_Laino

the requested equation is:
\[v = \sqrt {\frac{g}{{2 \times h}}} \times d = 4.95\;m/\sec \]

- Michele_Laino

|dw:1432834613891:dw|

- anonymous

ohhhhhh okay i think i get it, what about the second?

- Michele_Laino

please, I'm pondering...

- Michele_Laino

I think the first option, since we have to apply the third law of Newton, and the fourth option since for an isolate system the total momentum is conserved

- anonymous

its asking which ones are true and which are false, what are you saying those are?

- Michele_Laino

first option is true,
second and third options are false,
fourth option is true,
fifth option is false

- anonymous

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.

- Michele_Laino

here we have to apply the formula of Newton, about the interaction force between two object of masses M, and m respectively, which are separated by a distance d
\[F = G\frac{{Mm}}{{{d^2}}}\]

- anonymous

okay, so what goes where in the equation?

- Michele_Laino

the weight W of an object whose mass is m, on the earth surface is given by the subsequent formula:
\[\Large W = G\frac{{{M_T}m}}{{{R^2}}}\]
where R is the earth radius

- Michele_Laino

the weight of the same object, whose mass is m, at an altitude h with respect to the earth surface is given by the subsequent formula:
\[\Large {W_1} = G\frac{{{M_T}m}}{{{{\left( {R + h} \right)}^2}}}\]

- Michele_Laino

in both formulas M_T is the mass of the earth

- Michele_Laino

now, we have to use this condition:
\[\Large \frac{{{W_1}}}{W} = \frac{1}{{16}}\]

- Michele_Laino

after a substitution, we get this condition:
\[\Large {\left( {\frac{R}{{R + h}}} \right)^2} = \frac{1}{{16}}\]
what is h?

- anonymous

the height?

- Michele_Laino

yes!

- Michele_Laino

|dw:1432838068670:dw|

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