Physics, asked by niramalpradhan3478, 9 hours ago

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

Answers

Answered by vasquezlillian96
0

Answer:

R=

g

2V

o

2

cos

2

ϕ

sinθcos(θ+ϕ)

Explanation:

Given that

Speed of particle = Vo

Angle of particle from incline surface = θ

Angle of incline surface from horizontal =Φ

The formula for range R given as

If particle is moving upward

R=\dfrac{2V_o^2}{g}\dfrac{sin\theta cos(\theta +\phi ) }{cos^2\phi }R=

g

2V

o

2

cos

2

ϕ

sinθcos(θ+ϕ)

Time flight T

T=\dfrac{2V_osin\theta }{gcos\phi }T=

gcosϕ

2V

o

sinθ

Maximum height h

T=\dfrac{V_o^2sin^2\theta }{2gcos\phi }T=

2gcosϕ

V

o

2

sin

2

θ

Explanation:

hope it help

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