Time to Reach 12m: Solving the Physics Behind It

[leprofece]

A heel kick vertically upwards from the floor and a student in a window see it go up in front of it to 5.00 m/s. The window is 12.0 m above the floor. no taking into account the air resistance . How long it takes to reach that point?

The distance to reach is 1.27 m plus 12m

I Only want to know the answer the book used
t = sqrt ( ( 2.13,27)/9.8 )
But why not to use
y = 13.27+ 5t -5t²?

This is why i ask this question
can anybody help me??

 

[MarkFL]

I moved this thread as it is algebra based physics, but I am not sure how to give it a more descriptive title because I don't know what is being asked and the given answer from your textbook does not make sense to me.

t = sqrt ( ( 2.13,27)/9.8 )

What does that mean?

A diagram would be very helpful here to better convey what is being asked. :D

 

[zzephod]

A heel kick vertically upwards from the floor and a student in a window see it go up in front of it to 5.00 m/s. The window is 12.0 m above the floor. no taking into account the air resistance . How long it takes to reach that point?

The distance to reach is 1.27 m plus 12m

I Only want to know the answer the book used
t = sqrt ( ( 2.13,27)/9.8 )
But why not to use
y = 13.27+ 5t -5t²?

This is why i ask this question
can anybody help me??

It is imposible to follow the English of your question exactly, but it is possible to comment on what you ask at the end.

1. First the question is using a value of $$9.8$$ for g not $$10$$.

2. Your expression $$y = 13.27+ 5t -5t^2$$ is for displacement not time which is what the question asks for.

3. The answer the book uses: $$t=\sqrt{(2\times 13.27)/9.8}$$, is the time to fall from rest to the ground, where it has speed $$-5.15$$ m/s $$\approx -5$$m/s. So is approximately the time it takes a ball with kicked from the ground upwards with initial speed $$5$$ m/s to rise to $$13.27$$ m.

.

 

[leprofece]

It is imposible to follow the English of your question exactly, but it is possible to comment on what you ask at the end.

1. First the question is using a value of $$9.8$$ for g not $$10$$.

2. Your expression $$y = 13.27+ 5t -5t^2$$ is for displacement not time which is what the question asks for.

3. The answer the book uses: $$t=\sqrt{(2\times 13.27)/9.8}$$, is the time to fall from rest to the ground, where it has speed $$-5.15$$ m/s $$\approx -5$$m/s. So is approximately the time it takes a ball with kicked from the ground upwards with initial speed $$5$$ m/s to rise to $$13.27$$ m.

.

my expression 13.27+ 5t -5t² = 0
I ask why I cannot calculate time from this expresion or solving this equation??

 

[DavidCampen]

my expression 13.27+ 5t -5t² = 0
I ask why I cannot calculate time from this expresion or solving this equation??

You can use this, it is the exact solution, except that acceleration is 9.8 meters per second not 10, but you have to solve the quadratic equation..

 

[leprofece]

You can use this, it is the exact solution, except that acceleration is 9.8 meters per second not 10, but you have to solve the quadratic equation..

BUT THE BOOK Givis as a solution t = sqrt((2d)/g)
What is the right answer then ? and why??
I would appreciate your help!
Maybe Vo = 0 but i think at that time it is not in free fall

 

[DavidCampen]

Yes, the book's solution does not match ours. Perhaps there is an error in the problem statement or our interpretation. An object launched upward at a constant velocity of 5 m/s in the Earth's gravitational field will never attain a height of 13.27 meters. So perhaps we need to work backwards from the book's answer to find the error in the problem statement or our interpretation of it.

 

[DavidCampen]

As ZZephod said, the book's answer is the time it would take for an object to fall 13.27 meters in the Earth's gravitational field; about 1.65 seconds. At the constant upward velocity of 5 meters per second, and ignoring the effect of gravity, the object will rise only 8.25 meters in 1.65 seconds. So, after launching an object upward at 5 meters/second, it will have reversed course and struck the ground in less than 1.65 seconds while never reaching a height of 13.27 meters.

Trying to work backward from the book answer I just cannot find a reasonable problem statement that would give that answer.