Question

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

Answer 1

Answer:

Answer:

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

Explanation:

We are given that

Initial velocity of particle$\bf :\longmapsto{v_0}$

Angle of particle with inclined surface$\bf :\longmapsto{\theta}$

Angle of inclined surface with horizontal$\bf :\longmapsto{\phi}$

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

Total angle of particle with horizontal $\tt :\longmapsto{\theta+\phi}$

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

If particle moving upward

Time of flight$\bf :\implies{\frac{2v_0sin\theta}{gcos\phi}}$

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