Projectile Motion Range (please Help)

In summary: Vf_y= -11sin(31)i dont know how to make an image but i have side lengths 5,6,7In summary, the conversation discussed the calculation of the maximum range of a stone thrown at an angle of 31.0 degrees with a speed of 11.0 m/s. The first attempt at solving the problem involved drawing a triangle and using various formulas to find the time and distance traveled, but the answer was incorrect. The conversation then delved into the correct equations to use, including using the vertical and horizontal velocity and displacement equations. The final answer for the maximum range is not provided, but the correct final velocity in the y direction is determined to be -11sin(31).
  • #1
Cole07
106
0
Question: How far will a stone travel over level ground if it is thrown upward at an angle of 31.0 degrees with respect to the horizontal and with a speed of 11.0m/s?


What is the maximum range that could be achieved with the same initial speed?


I have tried to solve this with no sucess what i have done is Drawn a triangle with a angle being 31.0 degrees the opposite side from the angle is Vy the adjacent side is Vx the hypothesis is 11.0m/s then i drew a chart with X and Y X side of the chart I have Vx= 9.428840308 a=0 then on the y side i have
Vy=5.665418824 a=-9.8 the Vf=0m/s I used the formula Vf=Vi+At and solved for time and got
0.578103962 then i used d=Vi*t+1/2At^2 and solved for D and got 1.637620735 for the first question (But was told this was not correct)

For the second question i doubled the time and multiplied by Vx D=9.428840308 (1.15620784) and got 10.90169909 but it was wrong !

Is there anyone who might be able to help me Please? :cry:
 
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  • #2
what is the final velocity in the Y direction? And why?

for your second question how do you maximize the distance you could travel? They gave you a hint : the SPEED is thesame but something else must be different. What must change?
and Why did you double time??
 
  • #3
GOod grief you don't believe in rounding numbers do you?

If you've calculated, as it appears, the time taken for the vertical velocity to reach 0M/s, then you've calculated the time for the stone to reach it's maximum height, - it still has some way to go before it hits the ground.

I would use here the equation x = V(i)t + 0.5 a t^2
with the vertical velocity and a = 9.8 but remembering that when the stone hits the ground its vertical displacement is zero because it started from ground level

That will give you a time, then use this with the horizontal velocity (constant) to calculate the horizontal distance.
 
  • #4
final velocity in y direction is 0 because at the apex it is not going anywhere
 
  • #5
But the apex is not the final velocity. It comes back down again.
 
  • #6
Cole07 said:
final velocity in y direction is 0 because at the apex it is not going anywhere

tahts fine

but is that the final velocity in the y direction for the WHOLE trip

after all you are trying to find the time it took for the ENTIRE journey
 
  • #7
I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)
 
  • #8
Cole07 said:
I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)

howd you get it?
 
  • #9
That's not what I got.

What to you get for the time?
 
  • #10
i took 5.665418824(0.578103962)+0.5(-9.8)(0.57810396)^2 and got 1.637600545
 
  • #11
rsk said:
That's not what I got.

What to you get for the time?

Did you use [tex]y(t) = 11\cdot\sin(31)t-\frac{1}{2}gt^2=0[/tex] to calculate the time?

If you didn't, do so. (Solve for t.) When you do so, plug that time into [tex]x(t)=11\cdot\cos(31)t[/tex] to get the total displacement.
 
  • #12
to get the time i used the formula Vf=Vi+at
Vf=0m/s
a= -9.8
Vi=5.665418824 (by 11.0sin(31.0))
and then i solved for time
 
  • #13
Cole07 said:
to get the time i used the formula Vf=Vi+at
Vf=0m/s
a= -9.8
Vi=5.665418824 (by 11.0sin(31.0))
and then i solved for time

why are you still saying velocity final in the Y is zero? Its NOT zero

after it has completed its entire flight what is the velocity in the Y just before it hits the ground? I certainly hope it isn't zero otherwise whatever I've been doing for these 6 odds years has been in vain! VAIN!

and this world has no justice :frown:
 
  • #14
I don't understand? I am going by the example we did in class and for Vf we used 0 and this is at the apex
 
  • #15
Cole07 said:
I don't understand? I am going by the example we did in class and for Vf we used 0 and this is at the apex

Interesting class, indeed.

Btw, I suggest you do a few searches named 'projectile motion', this could do good. :smile:
 
  • #16
That would be good to calculate the maximum HEIGHT of the object, not the maximum distance it travels.

Throw a ball into the air. When it hits the ground, is the y velocity zero? Use intuition, it really helps sometimes
 
  • #17
since I am such a nice guy i drew a diagram for you
look at the diagram

now tell me what do u think the velocity i nteh Y direction must be after thw whole trip?

THINK before you answer

but don't think too hard :wink:

you are probably mixing up the type of questions
 

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  • #18
yes you are a nice guy but not only am i having trouble with physics BUT I CAN'T OPEN THE ATTACHMENT THIS IS SO SAD!
 
  • #19
..since I'm an even nicer guy, I'll give you this Java applet to play with, and hopefully learn something from it: http://www.walter-fendt.de/ph11e/projectile.htm" [Broken].

{:cool:,:tongue2:}
 
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  • #20
Cole07 said:
SHE is not taking a piss or washing her face she is still to work this f-ing problem out the horizontial distance is 10.9 meters but i still don't understand how to get Max. range i thought i used D=vt but i guess not!

correction SHE oooops :biggrin:

k does the image work now?? look at the image. What is the final velocity in teh Y answer that question. THINK before you answer it

And tell me WHY WHY WHYYYYYY you said that
 
  • #21
Image does not work
 
  • #22
ok i still don't have an image so i am thinking Vf would be -11 right?
 
  • #23
Cole07 said:
ok i still don't have an image so i am thinking Vf would be -11 right?

FINALLY

yesssssssssssssssssssssssssssssss correct

since the ball started from some point and lands at the same height the speed is th same, just different direction

so the final velocity in teh Y direction is just the negative of the initial velocity
 
  • #24
could you give me a clue as to how to get the maximum range i still have no image and am clueless
and frustrated!
 
  • #25
I take this as everyone gave up on me since radou and stunner logged off anybody else out there can you give me some help?
 
  • #26
Cole07 said:
could you give me a clue as to how to get the maximum range i still have no image and am clueless
and frustrated!

im still there

well if u have a garden hose and u have water coming out of it there is s certain angle at which the water will have maximum range and that angle is

*drumroll*



45 degrees!

so from now on if you have projectile motion you angle should be 45 degrees if u want to maximize the range both horizontally and vertically.

You should remember this FOREVER

so if i were to ask you this in aclub while you were piss drunk you should still be able to tell me 45 degrees and then fall over
:wink:
 
  • #27
No offense, but the two of them (as well as everyone else here) are not here just for you. If you want 24/7 support, hire a tutor.

https://www.physicsforums.com/showthread.php?t=94379
 
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  • #28
As you know projectile motion can be separated into motion with constant speed along the x-axis and motion with constant acceleration downwards in the y-direction. If the projectile is launched with a speed V at an angle [tex]\alpha[/tex] with repect to the x-axis we have that its speed in the x-direction is given by

[tex]V\cos(\alpha)[/tex]

and since

[tex]V\cos(\alpha) = \frac{x}{t}[/tex]

we have that

[tex]t = \frac{x}{V\cos(\alpha)}[/tex] ... (1)

the launching speed in the y-direction is given by

[tex]V\sin(\alpha)[/tex]

therefore the y-coordinate is given by

[tex]y = Vt\sin(\alpha) - 0.5gt^2[/tex] ... (2)

substituting (1) into (2) gives the parabolic equation for the trajectory of the projectile

[tex]y = \tan(\alpha) - \frac{g}{2V^2\cos^2(\alpha)}x^2[/tex] ... (3)

when the projectile hits the ground its y-coordinate will be zero. Its range, R will then be according to (3)

[tex]R = \frac{2V^2\sin(\alpha)\cos(\alpha)}{g}[/tex]

which can be simplified to

[tex]R = \frac{V^2\sin(2\alpha)}{g}[/tex]

the range will be a maximum when the sin is equal to 1. This will happen when

[tex]2\alpha = 90^o[/tex]

or [tex]\alpha = 45^o[/tex]

so the projectile will obtain its maximum range when it is launched at 45 degrees in which case its range will be

[tex]R = \frac{V^2}{g}[/tex]
 
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1. What is projectile motion range?

Projectile motion range is the horizontal distance that a projectile travels before hitting the ground. It is affected by the initial velocity, angle of launch, and gravitational acceleration.

2. How is projectile motion range calculated?

The range of projectile motion can be calculated using the formula: R = (v02 * sin(2θ)) / g, where R is the range, v0 is the initial velocity, θ is the angle of launch, and g is the gravitational acceleration.

3. What factors affect the range of projectile motion?

The range of projectile motion is affected by the initial velocity, angle of launch, and gravitational acceleration. Other factors that may affect the range include air resistance, wind, and the shape or weight of the projectile.

4. How does changing the angle of launch affect the range of projectile motion?

The range of projectile motion is maximized when the angle of launch is 45 degrees. As the angle of launch increases or decreases from 45 degrees, the range will decrease. If the angle of launch is 0 degrees (launched horizontally) or 90 degrees (launched vertically), the range will be 0.

5. Can the range of projectile motion be greater than the initial velocity?

No, the range of projectile motion cannot be greater than the initial velocity. The maximum range is achieved when the angle of launch is 45 degrees, and the range is equal to (v02 / g). If the initial velocity is increased, the range will also increase proportionally.

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