Question

Solve the following problem.

A particle is projected with speed v₀ at angle θ to the horizontal on an inclined surface making an angle Φ (Φ < θ) to the horizontal. Find the range of the projectile along the inclined surface.

Answer 1

1. The equation of trajectory of projectile is given by,
   y = `(tan theta)"x" - ["g"/(2"u"^2cos^2theta)]"x"^2`   ...(1)
2. In this case to find R substitute,
   y = R sin Φ               ….(2)
   x = R cos Φ              ….(3)
3. From equations (1), (2) and (3), we have,
   R sin Φ = tan θ (R cos Φ) - `("g"/(2"v"_0^2 cos^2theta)) "R"^2cos^2phi`    ....(∵ u = v₀)
4. So, sin Φ = tan θ cos Φ - `("g R" cos^2 phi)/(2 "v"_0^2 cos^2 theta)`
   ∴ `("g R" cos^2 phi)/(2 "v"_0^2 cos^2 theta) = tan theta cos phi - sin phi`
5. Hence,
   R = `(2"v"_0^2)/"g" [(cos^2theta)/(cos^2phi)] [tan theta cos phi - sin phi]`
   `= (2"v"_0^2)/"g" (cos theta)/(cos^2phi) [cos theta (sin theta)/(cos theta) cos phi - cos theta sin phi]`
6. So, R = `(2"v"_0^2)/"g" = (cos theta)/(cos^2phi) [sin theta cos phi - cos theta sin phi]`
   ∴ R = `(2"v"_0^2)/"g" = (cos theta)/(cos^2phi) sin (theta - phi)     .....[because sin(theta - phi) = sin theta cos phi - cos theta sin phi]`