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A stone is projected vertically upwards

  • Thread starter tehmatriks
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  • #1
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Homework Statement


A stone is projected vertically upwards from the ground with an initial velocity of 80 m s-1. Find the greatest height reached and the time taken. When will the stone be 96 m above the ground?


Homework Equations


[v = u + at], [s = ut + ½at2], [v2 = u2 + 2as]

assume all falling objects have an acceleration of 9.8 m s-2 vertically downwards.

The Attempt at a Solution


accerleration "a" = -9.8 m s-2
initial velocity "u" = 80 m s-1
final velocity "v" = 0 m s-1
time "t" = ? s


finding time "t" using
v = u + at
0 = 80 + (-9.8)t
t = 80/9.8
t = 8.16 s

finding greatest height i.e distance "s" using
v2 = u2 + 2as
02 = 802 + 2(-9.8)s
s = 802/2(9.8)
s = 326.53

Now for the second bit, finding when the stone will be 96 m above the ground.
 

Answers and Replies

  • #2
verty
Homework Helper
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Everything you have done is correct. You now need to find the time given the height. Of course there will be two times because the stone will pass 96m on the way up and on the way down. Which formula will you use?
 
  • #3
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Everything you have done is correct. You now need to find the time given the height. Of course there will be two times because the stone will pass 96m on the way up and on the way down. Which formula will you use?
that's the problem i dont know, non look right
so i must be missing something, i just can't figure out what

it's funny cuz a couple of days ago i was flying through all these questions, i take a couple of days off and forget simple steps and now im stuck in what i feel is probabley gonna be an answer so simple im not gonna feel right unless i slap myself for being so thick.
 
  • #4
verty
Homework Helper
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198
Let's see. When acceleration is constant, you can use the three formulas above.

1. v = u + at
2. v^2 = u^2 + 2as
3. s = ut + (1/2) a t^2

Acceleration is constant in this question, it is just -9.8. You want to find time. So you need a formula that you know everything about except time.

Formula 2 doesn't count, it doesn't have time in it. So you must choose formula 1 or 3. Formula 1 doesn't have displacement, which is a problem because you are interested in the time to reach a displacement of 96m. So you will have to use formula 3.

So you start there, plug everything into formula 3 and then try to find the time. Please do that and get as far as you can before responding again.

PS. I assume you are in college or your final year of high school, if not then let me know.
 
Last edited:
  • #5
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96 = 80t + 4.9t2

no idea what to do now
 
  • #6
verty
Homework Helper
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198
Okay, so you've found a formula that has only one unknown, but it has the form:

[tex]at^2 + bt + c = 0[/tex] for some a,b,c. Equations of this form are called quadratic equations. To solve them, we have the quadratic formula,

[tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.[/tex]

This is a more complex formula than perhaps you've seen before, so I'll show you where it comes from. You won't need to do these steps but see if you can follow along.

[tex]4.9 t^2 + 80 t = 96[/tex]

[tex]t^2 + \frac{80}{4.9} t = \frac{96}{4.9}[/tex]

[tex]t^2 + 2 \frac{80}{9.8} t + (\frac{80}{9.8})^2 = \frac{96}{4.9} + (\frac{80}{9.8})^2[/tex]

[tex](t + \frac{80}{9.8})^2 = \frac{192}{9.8} + (\frac{80}{9.8})^2[/tex]

[tex](t + \frac{80}{9.8})^2 = \frac{192*9.8 + 80^2}{9.8^2}[/tex]

[tex]t + \frac{80}{9.8} = \frac{\pm \sqrt{80^2 + 192*9.8}}{9.8}[/tex]

[tex]t = -\frac{80}{9.8} \pm \frac{\sqrt{80^2 + (2*96)*(2*4.9)}}{9.8}[/tex]

[tex]t = \frac{-80 \pm \sqrt{80^2 + 4*4.9*96}}{9.8}[/tex]

[tex]t = \frac{-80 \pm \sqrt{80^2 - 4*4.9*(-96)}}{9.8}[/tex]

And you'll see that this is the same as the quadratic formula when we substitute the values for a,b,c. Take particular note that c = -96 then.

So okay, you won't need to do those steps. You would jump straight to the last line. Now there are two times at which the stone has a displacement of 96m, and we can see that this formula has two solutions because the square root can be negative or positive.

Ok, continue the process and write out neatly what the two solutions are.

The quadratic formula is important to know because it can save a lot of time and you will often have to solve quadratic equations. There was another way to solve this question but I wanted to be able to show you the quadratic equation because quadratic equations appear very often.
 
  • #7
40
0
Okay, so you've found a formula that has only one unknown, but it has the form:

[tex]at^2 + bt + c = 0[/tex] for some a,b,c. Equations of this form are called quadratic equations. To solve them, we have the quadratic formula,

[tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.[/tex]

This is a more complex formula than perhaps you've seen before, so I'll show you where it comes from. You won't need to do these steps but see if you can follow along.

[tex]4.9 t^2 + 80 t = 96[/tex]

[tex]t^2 + \frac{80}{4.9} t = \frac{96}{4.9}[/tex]

[tex]t^2 + 2 \frac{80}{9.8} t + (\frac{80}{9.8})^2 = \frac{96}{4.9} + (\frac{80}{9.8})^2[/tex]

[tex](t + \frac{80}{9.8})^2 = \frac{192}{9.8} + (\frac{80}{9.8})^2[/tex]

[tex](t + \frac{80}{9.8})^2 = \frac{192*9.8 + 80^2}{9.8^2}[/tex]

[tex]t + \frac{80}{9.8} = \frac{\pm \sqrt{80^2 + 192*9.8}}{9.8}[/tex]

[tex]t = -\frac{80}{9.8} \pm \frac{\sqrt{80^2 + (2*96)*(2*4.9)}}{9.8}[/tex]

[tex]t = \frac{-80 \pm \sqrt{80^2 + 4*4.9*96}}{9.8}[/tex]

[tex]t = \frac{-80 \pm \sqrt{80^2 - 4*4.9*(-96)}}{9.8}[/tex]

And you'll see that this is the same as the quadratic formula when we substitute the values for a,b,c. Take particular note that c = -96 then.

So okay, you won't need to do those steps. You would jump straight to the last line. Now there are two times at which the stone has a displacement of 96m, and we can see that this formula has two solutions because the square root can be negative or positive.

Ok, continue the process and write out neatly what the two solutions are.

The quadratic formula is important to know because it can save a lot of time and you will often have to solve quadratic equations. There was another way to solve this question but I wanted to be able to show you the quadratic equation because quadratic equations appear very often.
wow man, thanks alot, i never thought i'd be using the b formula(that's what the teacher calls it) in physics
yes im in highschool.
and should'nt it be a "+" for the time being in the last equation?

i tried using the latex image thing but i keep getting generation failed cuz im obviously doing something wrong, so i decided to go straight to the last bit or else it'd just be too messy and complicating. *i copied and pasted this bit from your post and put a + instead of minus*

[tex]t = \frac{-80 \pm \sqrt{80^2 + 4*4.9*(-96)}}{9.8}[/tex]

-80 +- 67.2/9.8

-80 + 67.2/9.8 = 1.306 s on the way up

-80 - 67.2/9.8 = 15.02 s im guessing on way down

so what was the other way of doing it?
btw, thnks for taking a bit of your time to help me out, just know that this is much appreciated. :biggrin:
 
Last edited:
  • #8
verty
Homework Helper
2,164
198
Oh, there was a slight mistake with the signs. Anyway, forget I said anything about any other method. There is no other method. This is the method that must be used to solve this problem. It took a bit of explaining because it is a difficult problem, one that needs the quadratic formula.

Thank you, I'm just glad it made sense. Just remember to use enough precision so you don't get rounding errors.
 

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