A particle starts from the origin of co-ordinates at time t=0 and moves in the xy plane with a constant acceleration k in the y-axis direction.Its equation of motion is y^2=px^2,where p is a constant.Then,its velocity component in the x-direction is:

A)variable

B)underroot(2k/p)

C)k/2p

D)underroot(k/2p)

ANS:D

Dear Sameer,
we know by eq. of trajectory=>
y=xtan(o) - x^2tan(o)/R
...Where R=range and (o)=angle of projection.....also here g=k
squaring both sides and comparing with the given eq.
We get ucos(o)=root(k/2p)ANS.

Regards.

the velocity component in x direction is variable not constant

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from the equation we can get the angle of projection or the tan of angle which will give cos Q

then the x component of velocity will be k/2p^1/2