Initial velocity in projectile motion

In summary, you are supposed to use g=9.81 m/sec and your measured values of s and h to find the initial velocity of a ball shot out of a gun. You were wondering what equation you use to find the initial velocity. You are getting there. We are now even getting a glimmer of the experiment you did!
  • #1
maverick99
74
0
I'm supposed to use g=9.81 m/sec and my measured values of s and h to find the initial velocity of a ball shot out of a gun i did in lab today. I was wondering what equation i use to find the initial velocity.
 
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  • #2
I'm sure you meant [itex] g=9.8 ms^{-2} [/itex].
This equation is the best:
[tex] \vec{v}(t)=\vec{v}_{0}+\vec{g}t [/tex]

Daniel.
 
  • #3
Could you explain what each of those letters represent?
 
  • #4
Perhaps one that relates the initial velocity to the the distance traveled and the time of flight? Seems like you need to examine the equations you have at your disposal. Maybe make and effort to look thorough your book and gather all of the motion equations together. Study the equations with emphasis on learning the meaning of each term.

Here is the most basic equation, from which all can be derived.

[tex] \ddot {x} = - g [/tex]

Where x increases up.

Did that help? Perhaps not, as you have not provided even so basic of information as to what level of course this is. Please provide more information. Could you show us the "candidate" equations? What are you given?
 
  • #5
Integral said:
Perhaps one that relates the initial velocity to the the distance traveled and the time of flight? Seems like you need to examine the equations you have at your disposal. Maybe make and effort to look thorough your book and gather all of the motion equations together. Study the equations with emphasis on learning the meaning of each term.

Here is the most basic equation, from which all can be derived.

[tex] \ddot {x} = - g [/tex]

Where x increases up.

Did that help? Perhaps not, as you have not provided even so basic of information as to what level of course this is. Please provide more information. Could you show us the "candidate" equations? What are you given?

Heres what the previous steps

Since the initial velocity is only in the x-direction, v=vox, is vox =s/t and t can be determined from the y equation since the y displacemnt is equal to the height and is due solely to the acceleration of gravity: t=square root of (2h/g). Use g=9.81m/sec and your measured values of s and h to find the initial velocity of the ball.

s= 284.5cm, h=76.5cm

I'm also given h=1/2gt(squared)
 
  • #6
You are getting there. We are now even getting a glimmer of the experiment you did!

Seems like you have the equation and the information requiered to compute the time.
t=square root of (2h/g).

Or in our Latex capable fourm

[tex] t = \sqrt { \frac {2h} g [/tex]

(click on the equation to see what I typed to get that)

Use that time to get your last result.

Edit: Whoops I just recalled that you need [itex] V_0 [/itex]

Surely you have some equations relating [itex] V_0 [/itex] to your other values? What is that equaition?
 
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  • #7
Integral said:
You are getting there. We are now even getting a glimmer of the experiment you did!

Seems like you have the equation and the information requiered to compute the time.

Or in our Latex capable fourm

[tex] t = \sqrt { \frac {2h} g [/tex]

(click on the equation to see what I typed to get that)

Use that time to get your last result.

Edit: Whoops I just recalled that you need [itex] V_0 [/itex]

Surely you have some equations relating [itex] V_0 [/itex] to your other values? What is that equaition?


I see 2 equations.

Vox = s/t
r = Vot + 1/2at(squared)
 
  • #8
Here's something else too

The initial velocity components are:
Vox = Vo(Cos angle) and Voy = Vo(Sin angle)
 
  • #9
Yeah,then we could say you got the picture right. :smile:

Daniel.
 
  • #10
maverick99 said:
Here's something else too

The initial velocity components are:
Vox = Vo(Cos angle) and Voy = Vo(Sin angle)
However, you said before that the INITIAL velocity had no vertical component and was just horizontal. Thus, Voy=0. Try to focus on your original set of data and equations: (your words are quoted below)

Since the initial velocity is only in the x-direction, v=vox, is vox =s/t and t can be determined from the y equation since the y displacemnt is equal to the height and is due solely to the acceleration of gravity: t=square root of (2h/g). Use g=9.81m/sec and your measured values of s and h to find the initial velocity of the ball.
s= 284.5cm, h=76.5cm
I'm also given h=1/2gt(squared)
t = sqrt(2*h/g)
Vox = s/t
Do you see the technique??
 
  • #11
xanthym said:
However, you said before that the INITIAL velocity had no vertical component and was just horizontal. Thus, Voy=0. Try to focus on your original set of data and equations: (your words are quoted below)


t = sqrt(2*h/g)
Vox = s/t
Do you see the technique??

So is Vox = s/t the equation i need to find the initial velocity?
 
  • #12
That equation requires values for "s" and "t".
You measured "s".
How would you determine "t" in that equation??
 
  • #13
xanthym said:
That equation requires values for "s" and "t".
You measured "s".
How would you determine "t" in that equation??


use the equation t=2*h/g?
 
  • #14
use the equation t=2*h/g?
CORRECT!
Now substitute your measured values in the appropriate equations, and solve for Vox. Give it a try!
(Show all your work.)
~
 
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  • #15
xanthym said:
CORRECT!
Now substitute your measured values in the appropriate equations, and solve for Vox. Give it a try!
(Show all your work.)
~

Vox = s/t
284.5cm/3.95s
i ended up getting 72, would the units be cm/sec?
 
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  • #16
maverick99 said:
i ended up getting 72, would the units be cm/sec?
Check your UNITS.
(Hint #1: Are ALL your calculation units in cm?)
(Hint #2: g)
 
  • #17
yah it's all in cm except for the gravity which is 9.81

I'm not sure how to put the gravity into it...
 
  • #18
The quickest method is to convert g to "cm/sec^2" from it current "m/sec^2":
g = (9.8 m/sec^2) = (980 cm/sec^2)
You need to repeat ALL calculations with everything the same EXCEPT now use g=(980) in the first equation. Please show all your work.
~
 
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  • #19
t = sqrt(2*h/g)
Vox = s/t

so, h=1/2gtsqred, so h=1/2(981cm/s)(3.95s)squared which is 7653

t=sqrt(2*7653/980cm/s) which is 3.95s
Vox = s/t so, 284.5cm/3.95s which finally turns out to be 72
is this right??
 
  • #20
maverick99 said:
t = sqrt(2*h/g)
Vox = s/t

so, h=1/2gtsqred, so h=1/2(981cm/s)(3.95s)squared which is 7653

t=sqrt(2*7653/980cm/s) which is 3.95s
Vox = s/t so, 284.5cm/3.95s which finally turns out to be 72
is this right??
NO. It is not correct this time.
You originally performed the calculations correctly except that "g" had the wrong units. Let's return to your original calculation technique:

THE EQUATIONS:
Equation #1 -----> t = sqrt(2*h/g)
Equation #2 -----> Vox = s/t

YOUR MEASURED VALUES:
s= 284.5cm, h=76.5cm

STEP #1:
Calculate "t" from Eq #1 using your MEASURED value of h=(76.5cm) and the known constant g=(980 cm/sec^2)

STEP #2:
Calculate "Vox" from Eq #2 using your MEASURED value of s=(284.5cm) and the value of "t" calculated from Step #1.

You are very close to the correct answer. Take your time and focus on the steps above.

~~
 
Last edited:
  • #21
xanthym said:
NO. It is not correct this time.
You originally performed the calculations correctly except that "g" had the wrong units. Let's return to your original calculation technique:

THE EQUATIONS:
Equation #1 -----> t = sqrt(2*h/g)
Equation #2 -----> Vox = s/t

YOUR MEASURED VALUES:
s= 284.5cm, h=76.5cm

STEP #1:
Calculate "t" from Eq #1 using your MEASURED value of h=(76.5cm) and the known constant g=(980 cm/sec^2)

STEP #2:
Calculate "Vox" from Eq #2 using your MEASURED value of s=(284.5cm) and the value of "t" calculated from Step #1.

You are very close to the correct answer. Take your time and focus on the steps above.

~~

k i did the steps and got 720, is that right? if so, 720 what? i don't know what the units would be
 
  • #22
maverick99 said:
k i did the steps and got 720, is that right? if so, 720 what? i don't know what the units would be
CORRECT!
Congratulations!
The initial velocity is (720 cm/sec) since all your units were in "cm" and "sec".


~~
 
  • #23
xanthym said:
CORRECT!
Congratulations!
The initial velocity is (720 cm/sec) since all your units were in "cm" and "sec".


~~
alright cool, thanks for the help!
 

What is initial velocity in projectile motion?

Initial velocity in projectile motion refers to the speed and direction at which an object is launched or thrown into the air. It is the first velocity that the object has as it begins its motion.

How is initial velocity calculated?

Initial velocity can be calculated by dividing the horizontal distance traveled by the time it took to travel that distance. This can be represented by the equation v = d/t, where v is initial velocity, d is the distance, and t is the time.

Why is initial velocity important in projectile motion?

Initial velocity is important because it determines the path and trajectory of an object in projectile motion. The direction and speed at which an object is launched will determine its motion and eventual landing point.

Does initial velocity affect the height of a projectile?

Yes, initial velocity does affect the height of a projectile. A higher initial velocity will result in a higher trajectory and therefore a higher maximum height for the projectile.

How does initial velocity change during projectile motion?

Initial velocity does not change during projectile motion unless there is a force acting on the object. In an ideal scenario with no external forces, the initial velocity will remain constant throughout the motion of the projectile.

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