IIT JEE , AIEEE Forum - Physics , Mechanics - vaibhav mechanics [Admin]: cartesian equation of trajectory

vaibhav1729

a projectile is given an initial velocity i+2jthe cartesian equation of its trajectory when g=10 is

gorakavipraveen

Its not a big issue to understand vibhav. See it does mean that v=root5 and angle tan^(-1)2 ie taninvere 2. so you simply apply it with substituting time as u/cos@ @thete in vertical displacement law. This give you after substituting the above values ,
y=2x-5x^2.
this is the reuqired parabola
bye

kirtana

hii
the eqution of a trajectory is given by,
y = x tan @ - gx² / 2u²cos² @

now,
u cos @ = 1
and u sin @ = 2

so.. tan @ = 2
and now substitute..
so..
y = 2 x - 10 x² / 2* 1
therefore,
y= 2x - 5 x²

innit right??

rashmi_jain

U_x=1

U_y=2

x=U_x * t ----------- eqn (1)

y=U_y*t-0.5*g*t² ----------- eqn(2)

from eqn(1)

x=t

y=2t-5t²

therefore, y=2x-5x²

cvramana

The magnitude of velocity of projection is given by u = Ö(1² + 2²) = Ö5 m/s. And the angle of projection is given by q where, tan q = 2 /1 = 2.

Therefore

Y = x tan q - g x² (1 + tan² q) / 2 u².

Y= 2x - 5x².

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