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Question about Projectiles being launched horizontally

  • Thread starter Sonny18n
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  • #1
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Homework Statement



Stone is thrown at +15 m/s from top of cliff at 44 meters high.
Im looking for the time it takes to reach the bottom and the final answer is 3 seconds. But I dont know how to get there.
Equation starts with
t^2 =2Y/g Y is initial height meaning -44 meters. The g is -9.80m/s^2. But where does the 2 come from? Is it always there in the formula?


(Sorry for the format, hard to copy and paste on phone)
 

Answers and Replies

  • #2
berkeman
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Homework Statement



Stone is thrown at +15 m/s from top of cliff at 44 meters high.
Im looking for the time it takes to reach the bottom and the final answer is 3 seconds. But I dont know how to get there.
Equation starts with
t^2 =2Y/g Y is initial height meaning -44 meters. The g is -9.80m/s^2. But where does the 2 come from? Is it always there in the formula?


(Sorry for the format, hard to copy and paste on phone)
What direction is the stone thrown? The initial vertical velocity will change how long it takes to hit bottom, right?

And can you list each of the kinematic equations of motion for reference please? The one for position as a function of time is particularly important (and it relates to the equation you show)...
 
  • #3
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What direction is the stone thrown? The initial vertical velocity will change how long it takes to hit bottom, right?

And can you list each of the kinematic equations of motion for reference please? The one for position as a function of time is particularly important (and it relates to the equation you show)...
It only says it's thrown horizontally. The only equations I have is x=Vxt and y=Vyt + 1/2gt^2
 
  • #4
berkeman
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It only says it's thrown horizontally. The only equations I have is x=Vxt and y=Vyt + 1/2gt^2
Ah, thrown "horizontally"! You left that out of your original post (OP). :-)

And the 2nd equation you list in the quoted text is incomplete. Please double check what the full equation is for the vertical position in projectile motion for y(t) as a function of initial y position, initial y velocity, and constant acceleration in the y direction...
 
  • #5
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Ah, thrown "horizontally"! You left that out of your original post (OP). :)

And the 2nd equation you list in the quoted text is incomplete. Please double check what the full equation is for the vertical position in projectile motion for y(t) as a function of initial y position, initial y velocity, and constant acceleration in the y direction...
That's all I have I think. Maybe this'll help:
Y=vyt + 1/2gt^2, or with vy=0, y = 1/2gt^2. Thus t^2 = 2y/g
 
  • #6
berkeman
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That's all I have I think. Maybe this'll help:
Y=vyt + 1/2gt^2, or with vy=0, y = 1/2gt^2. Thus t^2 = 2y/g
That equation doesn't include a starting y position though, does it? Your textbook should list all of the kinematic equations. If not, just use wikipedia to get the full form of the position equation for constant acceleration. BTW Quiz Question -- why is the acceleration constant in this problem?
 
  • #7
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If by y
That equation doesn't include a starting y position though, does it? Your textbook should list all of the kinematic equations. If not, just use wikipedia to get the full form of the position equation for constant acceleration. BTW Quiz Question -- why is the acceleration constant in this problem?
is the y position the initial vertical velocity? If so, then its zero.

To your question: I've never seem that term
 
  • #8
berkeman
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If by y

is the y position the initial vertical velocity? If so, then its zero.

To your question: I've never seem that term
It's important in this problem to use the full equation for vertical position as a function of time given a constant vertical acceleration. Please check wikipedia or some other source to get the equation. And please answer my Quiz Question -- it will help you get a more intuitive understanding of what is going on in these projectile motion problems. :)
 
  • #9
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It's important in this problem to use the full equation for vertical position as a function of time given a constant vertical acceleration. Please check wikipedia or some other source to get the equation. And please answer my Quiz Question -- it will help you get a more intuitive understanding of what is going on in these projectile motion problems. :)
Is the wikipedia page called "free fall"
 
  • #10
berkeman
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Is the wikipedia page called "free fall"
I would go to the wikipedia web page and search on Kinematic Equations of Motion. IIRC, that entry starts pretty complicated, then simplifies to the equations for constant acceleration...
 
  • #11
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I would go to the wikipedia web page and search on Kinematic Equations of Motion. IIRC, that entry starts pretty complicated, then simplifies to the equations for constant acceleration...
Wait, does this make sense?
t^2 = 2y/g=9.0
=2(-44m)/-9.80m/s^2= 9.0 s^2
=sqrt 9.0s^2 = 3.0s
 
  • #12
berkeman
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Wait, does this make sense?
t^2 = 2y/g=9.0
=2(-44m)/-9.80m/s^2= 9.0 s^2
=sqrt 9.0s^2 = 3.0s
It may be right or wrong, but I have no idea what it is.

There is a kinematic equation of motion for vertical motion given a constant acceleration. It involves y(t) = and 3 terms on the right hand side (RHS). Please find that equation and plug in the values for this problem. It is always best to start with the full equations when learning the basics, IMO.
 
  • #13
haruspex
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Wait, does this make sense?
t^2 = 2y/g=9.0
=2(-44m)/-9.80m/s^2= 9.0 s^2
=sqrt 9.0s^2 = 3.0s
Almost.
You should delete the "=9.0" in the first line. We don't know this yet.
And it's not correct to write
= 9.0 s^2
=sqrt 9.0s^2​
You should write
t^2 = 9.0 s^2
t = sqrt (9.0s^2)​
 
  • #14
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Almost.
You should delete the "=9.0" in the first line. We don't know this yet.
And it's not correct to write
= 9.0 s^2
=sqrt 9.0s^2​
You should write
t^2 = 9.0 s^2
t = sqrt (9.0s^2)​
Ok but do you know how the two is supposed to get there in the first line?
 
  • #15
haruspex
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Ok but do you know how the two is supposed to get there in the first line?
Consider an object accelerating uniformly at rate a from rest.
After time t is moving with velocity at.
Since it started from rest, its average velocity over time t was (at+0)/2 = at/2.
If a body has average velocity at/2 for time t it travels (at/2)*t = at2/2.
 
  • #16
Delphi51
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There is a reason why the high school physics demo of shooting a steel ball horizontally at a falling light bulb is enjoyed and remembered by most students. As long as the ball in 2D motion and the bulb in 1D motion start simultaneously, the ball always hits the bulb no matter what speed it is shot at. The vertical and horizontal motions of the ball are independent. You may find the time to fall of the object launched horizontally without considering its horizontal motion. Just vertical motion accelerated by gravity. A neat way to simplify any vertical and horizontal motion problem is to do the vertical and horizontal parts separately.
 

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