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Projectile motion and rangeby leah3000
Tags: projectile motion 
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#1
Mar710, 05:05 PM

P: 43

referring to an object being projected upwards at an angle ө
i'm a phys student but i don't do math so i'm having trouble understanding how this equation has been simplified. Horizontal Range= vcosө x 2vsinө/g =v^2 2sinөcosө/g this is the line i don't understand : v^2 sin^2 ө/g where did the cosө go? is is a trig identity that cancels? also when considering upward motion is g taken as negative? if so, then why is the time taken to reach the maximum pt given by: t= v sinө/ g wouldn't it be v sinө/ g isn't that the reason for the vertical distance travelled being given by; y= v sinө t 1/2 gt^2 ?? 


#2
Mar810, 12:27 AM

P: 647

Your intuition serves you well. I do think it was a trig identity that simplified your equation.
[tex]\sin\left( 2\theta\right)=2\sin\theta\cos\theta[/tex] also, you are right again kind of... You can make your coordinate system any way you want, just as long as you're consistent. So make gravity negative but that formula you gave for time comes from this formula [tex]v_{final}=at+v_{initial}[/tex] and then setting the final velocity to zero, and inserting negative gravity [tex]v_{initial}=gt[/tex] and solving for time [tex]\frac{v_{initial}}{g}=t[/tex] note here in my work that [tex]v\sin\theta = v[/tex] So there gravity is positive, but that is because the negative signs canceled out left and right. This should get you your answers. So gravity can be negative, you just have to make sure to stay consistent once you decide how you want to orient yourself. 


#3
Mar810, 02:18 PM

P: 43

wow...thank you so much. It really did clear things up. I've never seen that identity before though...i have very limited knowledge of basic trigs lol
so then i should be specifying whether i use gravity as +ve/ ve in general parabolic calculations? 


#4
Mar810, 04:09 PM

P: 647

Projectile motion and range



#5
Mar810, 08:07 PM

P: 4

i dont understand about why ax=o?



#6
Mar810, 08:37 PM

P: 43




#7
Mar810, 08:39 PM

P: 647

for most of these projectile problems, the acceleration [tex]a[/tex] and velocity [tex]v[/tex] can be broken down into their two x and y directions respectively [tex]a_{x}=a\cos\theta[/tex] and [tex]a_{y}=a\sin\theta[/tex] and [tex]v_{x}=v\cos\theta[/tex] and [tex]v_{y}=v\sin\theta[/tex] These would comprise the velocity and accelerations found in almost any arbitrary projectile question, and the ability to break down a vector into its components is key for solving these problems. So any problem could have an acceleration in the x direction, but i dont believe this is one of those cases, i think the projectile leaves at its trajectory at a constant [tex]v_{x}[/tex]. 


#8
Mar1310, 11:29 AM

P: 4

3 . A particle starts from the origin at t=0 with an initial velocity having an xcomponent of 20m/s and a ycomponent of 15m/s. The particle moves in the xy plane with an x component of acceleration only, given by ax = 4.0m/s2
a)Determine the total velocity vector at any time b)Calculate the velocity and speed of the particle at t=0.5s c)Determine the x and y coordinates of the particle at any time t and its position vector at this time what i know ax=o? so i m confused? 


#9
Mar1310, 02:06 PM

P: 647

I think the best approach for this problem would be to consider your velocity in terms of a vector and insert your acceleration vector into a kinematic equation. [tex]v_{f}=at+v_{i}[/tex] with your acceleration being [tex]a=4.0 \hat{i}[/tex] and your velocity being [tex]v_{i}=\left(20\hat{i}15\hat{j}\right)[/tex]. putting these into the kinematic equation would give your total velocity for any time. for the second part it is a matter of plugging in the time. for the third part remember that the integral of velocity is position, so you can integrate your equation wrt time and solve for position. this should point you in the right direction :) 


#10
Mar1310, 11:26 PM

P: 4

sorry for trouble you...:) .can you show the solution for all qeustion...because i dnt no so solve it...i very hope for you..



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