Projectile Motion Help: Proving Equations for Velocity and Time

AI Thread Summary
The discussion focuses on proving equations related to projectile motion, specifically involving velocities at three points along a projectile's path and the corresponding times taken to traverse segments of the path. The user has established relationships between the horizontal components of velocity and attempted to derive the required equations using trigonometric manipulations. However, they express difficulty in progressing further and question whether they are missing additional equations or if their trigonometric approach is flawed. The user has derived a secondary relation but struggles to find physical significance in the results. The conversation highlights the complexities of applying trigonometry to solve projectile motion problems effectively.
elphin
Messages
18
Reaction score
0
projectile motion help needed!

Homework Statement



If v1, v2, v3 are the velocities at three points A, B , C of the path of a projectile, where the inclinations to the horizon are α, α-β, α-2β and if t1, t2 are the times of describing the arcs AB and BC respectively, prove that

v3*t1 = v1*t2 and 1/ v1 + 1/v3 = (2*cos β)/v2


Homework Equations





The Attempt at a Solution



v1cos α = v2cos (α-β)= v3cos (α-2β) - (horizontal component same always)

and

v2sin(α-β) = v1sin α – g*t1 (&) v3sin (α-2β) = v2sin(α-β) – g*t2

and now I am stuck .. are there any more equations that I am missing or is it a bad case of trigonometric manipulation.. I tried trigonometric manipulation .. but ended up getting weirder equations
 
Physics news on Phys.org


I think it's a case of choosing the right trigonometric manipulations. I got the second relation-

Using your formulas,

\frac{1}{v_{1}}+\frac{1}{v_{2}}=\frac{cos(\alpha)}{v_{2} cos(\alpha-2\beta)}+\frac{cos(\alpha-\beta)}{v_{2} cos(\alpha-\beta)}

cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha) + cos(\alpha)cos(2\beta) - sin(\alpha)sin(2\beta)
cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha)(1+cos(2\beta)) - 2sin(\alpha)sin(\beta)cos(\beta)
cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha)(1+2cos^{2}(\beta)-1) - 2sin(\alpha)sin(\beta)cos(\beta)
cos(\alpha) + cos(\alpha-2\beta) = 2cos(\beta)(cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)) = 2cos(\beta)cos(\alpha-\beta)

Resubstitute and get the required result. However I don't see any physical significance to this..
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top