Question

wi) A particle is projected with speed v, at
angle o to the horizontal on an inclined
surface making an angle • (ø <0) to the
horizontal. Find the range of the projectile
along the inclined surface.

How to solve this?

Answer 1

Time taken by particle

From A to B is

t=2v[sin(tetta-tetta')]/g cos(tetta')

In this time period velocity component parallel to incline cover displacement is range.

Range(R)=v cos(tetta-alpha)t-1/2gsin(tetta)'t

=v cos(tetta-alpha)*2v(sin(tetta-tetta')/gcos tetta'-1/2g sin(tetta')'{2v/

=v2sin2(tetta-tetta')/g cos(tetta)'-2v2sin2(tetta-tetta')/g cos(tetta)'

=2v2sin (tetta-tetta')/g cos(tetta'){cos(tetta-tetta')-sin(tetta-tetta')}

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