How Does Projectile Angle Affect Range in Motion Physics?

[Cole07]

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?

 

[stunner5000pt]

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??

 

[rsk]

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.

 

[Cole07]

final velocity in y direction is 0 because at the apex it is not going anywhere

 

[rsk]

But the apex is not the final velocity. It comes back down again.

 

[stunner5000pt]

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

 

[Cole07]

I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)

 

[stunner5000pt]

I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)

howd you get it?

 

[rsk]

That's not what I got.

What to you get for the time?

 

[Cole07]

i took 5.665418824(0.578103962)+0.5(-9.8)(0.57810396)^2 and got 1.637600545

 

[radou]

That's not what I got.

What to you get for the time?

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

If you didn't, do so. (Solve for t.) When you do so, plug that time into x(t)=11\cdot\cos(31)t to get the total displacement.

 

[Cole07]

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

 

[stunner5000pt]

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

 

[Cole07]

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

 

[radou]

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.

 

[Office_Shredder]

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

 

[stunner5000pt]

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

you are probably mixing up the type of questions

 

[Cole07]

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!

 

[radou]

..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" .

{,}

 

[stunner5000pt]

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

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

 

[Cole07]

Image does not work

 

[Cole07]

ok i still don't have an image so i am thinking Vf would be -11 right?

 

[stunner5000pt]

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

 

[Cole07]

could you give me a clue as to how to get the maximum range i still have no image and am clueless
and frustrated!

 

[Cole07]

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?

 

[stunner5000pt]

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

 

[geoffjb]

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

 

[andrevdh]

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 \alpha with repect to the x-axis we have that its speed in the x-direction is given by

V\cos(\alpha)

and since

V\cos(\alpha) = \frac{x}{t}

we have that

t = \frac{x}{V\cos(\alpha)} ... (1)

the launching speed in the y-direction is given by

V\sin(\alpha)

therefore the y-coordinate is given by

y = Vt\sin(\alpha) - 0.5gt^2 ... (2)

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

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

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

R = \frac{2V^2\sin(\alpha)\cos(\alpha)}{g}

which can be simplified to

R = \frac{V^2\sin(2\alpha)}{g}

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

2\alpha = 90^o

or \alpha = 45^o

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

R = \frac{V^2}{g}