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Question

A particle is projected with speed Vo at an angle θ to the horizontal on an inclined surface making an angle ϕ (ϕ<θ) to the horizontal. Find an expression for its range along the inclined surface.

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Solution

uX=v0cos(θϕ)
uY=v0sin(θϕ)
aX=gsinϕ
aY=gcosϕ
R=uXt12gsinϕt2
vY0
uY+aYt=0
v0sin(θϕ)-gcosϕt=0
t=v0sin(θϕ)gcosϕ
Time of flight =T=2v0sin(θϕ)gcosϕ
R=uXT12gsinϕT2
=T(v0cos(θϕ)12gsinϕT)
=T(v0cos(θϕ)gsinϕ2×2v0sin(θϕ)gcosϕ)
=Tv0(cos(θϕ)sinϕsin(θϕ)cosϕ)
=v0T(cos(θϕ)cosϕsinϕsin(θϕ)cosϕ
=v0Tcosθcosϕ
=v0cosθcosϕ×2v0sin(θϕ)gcosϕ
R=2v20sin(θϕ)cosθgcos2ϕ

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