IIT JEE , AIEEE Forum - Physics , Mechanics

russs

a body is thrown from the surface of earth with an angle

to he horizontal.assuming the air drag to be negligible find the curvature radii of trajectory at its initial point and at its peak

ashish_banga

at the initial point it is u^2/ gcos alpha
at the peak =
u^2 cos^2 alpha / g

me_living4u

at initial point....

velo. = u

and acc. radial.. centripetal= u^2/R

this acc. is provided by acc. due to grav. = gcos alpha (see acc. only perp. to vel.)

so gcos alpha = u^2/R

=> R= u^2/gcos alpha

similarly at topmost point

acc. due to grav=g

and vel. = ucos alpha

so R= u^2cos^2 alpha/g

Cheers!!!!!

iamtop1

horizontal
ux= U Cos ?
motion along horizontal
x= Ux t +1/2 axt2
ax=0
x= U Cos ? t or t= x /U Cos ?

motion along vertical
y= Uy t + ½ ay t2

y= U sin ? t + ½ (-g) t2
= U sin ? t ? ½ g t2
Putting t = x/U Cos ?
y= U sin ? x/ U Cos ? - ½ g x2/U2 cos2 ?
tan ? - (g/2U2Cos2 ?) x2 called trajectory

sarthak321

A point undergoing any type of motion can assumed to be undergoing circular motion at any pt of its trajectory however the circles keep on changing with the position.

As we all know the centripital acceleration of a body is given by

²R
or simply V²net/R (as

R=V)

now at the highest pt of the trajectory the particle has net velocity in x direction

since in x we dont have any acceleration u remains constant and equal to ucos

now (ucos

)²/R=g (since centripital acceleration is perpendicular to Vnet and here g is perpendicular to ucos

at highest point))

so R=(ucos

)²/g

....
I hope u like the answer.....

as for the 1st part i.e initial veleocity take component of g perpendicular to u
and use it as R...

rate me if my answer helped

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