Question

A particle is projected with speed v0 at angle theta to the horizontal on an inclined surface making an angle phi(phi less than theta) to the horizontal. Find the range of the projectile along the inclined surface.

Answer 1

Answer:

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

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

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

Time flight T

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

Maximum height h

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

Answer 2

Answer:

$Range=\frac{2v^2_0sin\theta cos(\theta+\phi)}{gcos^2\phi}$

Explanation:

We are given that

Initial velocity of particle=$v_0$

Angle of particle with inclined surface=tex]\theta[/tex]

Angle of inclined surface with horizontal=$\phi$

We have to find the range of projectile along the inclined surface.

Total angle of particle with horizontal =$\theta+\phi$

Range:Total distance covered by particle in horizontal direction is called range.

If particle moving upward

Time of flight=$\frac{2v_0sin\theta}{gcos\phi}$

$Range=\frac{2v^2_0sin\theta cos(\theta+\phi)}{gcos^2\phi}$