Projectile and Horizontal Distance

In summary, the problem involves a projectile launched at an angle theta to the horizontal that reaches a maximum height h. The horizontal range of the projectile can be calculated using the equation 4h/tan(theta). To solve this problem, one must find the vertical and horizontal components of the initial velocity, calculate the maximum height reached using the vertical component as the initial velocity, and find the total time of flight until the vertical distance is zero again. Using these values, the horizontal distance traveled can be calculated.
  • #1
nightshade123
82
0
[SOLVED] Projectile and Horizontal Distance

Homework Statement



A projectile launched at an angle [tex]\theta[/tex] to the horizontal reaches a maximum height h. Show that its horizontal range is 4h/tan[tex]\theta[/tex]

Homework Equations



Can you help me figure out how to get this setup properly and what equations i should use?

The Attempt at a Solution



I can not figure out how to answer this question i have tried deriving my own forumla and none of them are coming out correctly
 
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  • #2
Find the vert and horz components of the initial velo. Then find max height reached using vert comp as the initial velo upward. That'll give you a relation between h and theta. Now find the total time of flight until vert dist is zero again, and from that the horz dist traveled in that time.

If you don't go through your books or notes, and write what you have done next time, it'll be difficult for us to help you.
 
  • #3
i have been going through my notes and online homework problems and looking online and reading the book and looking at his lecture slide shows, which blow... he doesn't teach we have to teach ourselves and work in groups.. my group couldn't figure it out either,

i don't get how i am soposed to get the vertical and horiz component

i got for the first part of this solution

tan theta = h / q if q if the horizontal axis

i have equation i could use but don't really see how to apply them

vector v = vector v not + vector a*t
vector r = vector r not + vector v not*t + 1/2 vector a*t^2

v sub x = v sub x not
v sub y = v sub y not - g * t
x = x not + v sub x not * t
y= y not + v sub y not * t - 1/2 * g * t^2projectile trajectory
y = x tan theta - g / ( 2 *v not ^2 * cos ^2 theta not ) * x^2

horizontal range
x = v not ^2 * sin 2 theta not / gtheta = tan^-1 A sub y / A sub x
A sub x = A cos theta
A sub y = A sin theta

my last problem left due for next week and it's killing me! lol
 
  • #4
(Post above deleted by OP.)
---------------------------

(You need not write v not, but write v0. The not is not "not" here, but nought. :smile:)

The problem with you is that that you know nought about projectile motion. It's not possible for us to write down the whole chapter explaining the concepts, or even write each of the eqns. Can you do some more studies with solved examples on your own?

Let's see what you can do in a couple of days, and then come back here. Next week is far away. :cool:
 
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  • #5
next week as in Monday..
 
  • #6
Do you know the eqns of st line motion under uniform accn, like vertical motion under gravity?

Do you know how far a particle goes in time t if projected witha velo v in absence of gravity?
 
  • #7
yes... of course...
 
  • #8
How much would a body rise up vertically if thrown up with a speed of 'u'? How much time would it take? What's the speed up at that highest point?

What's the answer to my 2nd Q? I wanted you to answer with the answer.

(There was no reason to delete your original post after post #2! At least that showed that you've gone through some mathematics.)
 
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  • #9
well if your just going to assume I am not trying to solve this problem and make fun of some information i gathered considering I am only a week or so into physics so i don't have equations and equation names memorized yet... i have talked to half a dozen people that havnt figured this problem out..

these are the equations i believe i can use to solve the problem...

accel is vertical and const downward
x=x0 + v0x t
vx = v0x

vertical motion has grav. accel. vertical motion is free fall
y = y0 + v0y t - 1/2 g t^2
vy = v0y - g t

trigonometric relations between components
v0x = v0 costheta
v0y= v0 sin thetaymax = (v0^2 * sintheta^2) / (2 * g)

ugh this is insane trying to achieve this relation
 
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  • #10
nightshade123 said:
well if your just going to assume I am not trying to solve this problem and make fun of some information i gathered considering I am only a week or so into physics so i don't have equations and equation names memorized yet... i have talked to half a dozen people that havnt figured this problem out..

On the contrary, you are going to need answers to exactly these questions to solve this problem. So you may understand that I had no intention of making fun of you. BTW, who were those half a dozen people?

(I’m writing in shorthand.)

The horz comp of the velo remains const = v*cos(theta). The vert comp of the velo undergoes the change.

Initial vert comp of velo = v*sin(theta)

If h is the max height, then 0 = (vsin(theta))^2 – 2gh (1)

If it takes time T to reach max height, then,
0 = vsin(theta) – gT => T = vsin(theta)/g. –(2)

The total time taken to fall back to ground again is 2T.

Horz dist covered in time 2T
= vcos(theta)*2T
= vcos(theta)*2vsin(theta)/g [from (2)]
= 4h/tan(theta) [using value of v^2 from (1)].
 
  • #11
i actualy had equation one, and i couldn't find how to relate time... i talk to my physics group and part of another group, and they said they got something and it didnt make sense or they were lost

i understand everything you did completely and it makes perfect sense, thanks for the help

i just hate people that ask me for help when they don't do ****, so i never ask for help until i have tried everything and put in a lot of time, i had a calc friend who would always come to me complaining he couldn't figure something out, turns out he put like 10 mins into it and gave up, i have prolly put about 10+ hours into this problem lol ugh.. i don't like to rely on people, even though i post on here I am not waiting for a response I am still working looking through all my crud on my desk

i know you got to be the same way you probely get a lot of plp that post w/o putting a lot of effort into solving it on there own, i kinda take it personally when someone tells me to do more when i know i have done everything that is available to me
 
  • #12
nightshade123 said:
i kinda take it personally when someone tells me to do more when i know i have done everything that is available to me
He's not breaking your balls. Look at your initial post. Does it look like it was posted by someone who has worked on it for 10+ hrs? Maybe you should spend 9 hrs doing the problem and another hour posting everything you've done. Don't get too revved up, you set yourself up.
 
  • #13
nightshade123 said:
i guess that's true my first time posting though, i wasnt sure how in depth i should have gone in what i had tried, i might try to scan my papers and post those
Yeah do that next time! A lot of people here are obviously smart and by just analyzing your work, they can spot the mistake w/o having to go thru the calculations themselves.
 
  • #14
A number of intermediate posts deleted by the OP. The OP should be warned against doing this.
 

Related to Projectile and Horizontal Distance

1. What is a projectile?

A projectile is an object that is launched or thrown into the air and then moves under the influence of gravity alone.

2. How does horizontal distance affect projectile motion?

Horizontal distance does not affect projectile motion as the horizontal component of motion is constant and does not change due to gravity.

3. What factors affect the horizontal distance of a projectile?

The horizontal distance of a projectile is affected by the initial velocity, launch angle, and air resistance. These factors can change the time of flight and the speed of the projectile, resulting in different horizontal distances.

4. What is the formula for calculating horizontal distance?

The formula for calculating horizontal distance is: Horizontal distance = initial velocity x time of flight x cosine of launch angle.

5. How can horizontal distance be increased in projectile motion?

Horizontal distance can be increased in projectile motion by increasing the initial velocity, launching the projectile at a higher angle, or reducing air resistance. These changes will result in a longer time of flight and a greater horizontal distance traveled.

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