Projectile Motion Q (1 Viewer)

[lyounamu]

A body is projected from a point on a horizontal plane reaches a greatest height h before hitting the plane R metres away from its point of projection. Show that the equation of the body's path may be expressed by y = [4hx(R-x)] / R^2

Sorry to bother you again. Thanks in advance.

 

[3unitz]

equation of motion: y = - kx(x - R) ------(1)

sub (R/2 , h) to find k:
h = - kR/2 (R/2 - R)
4h/R^2 = k

sub k into (1):
y = - (4h/R^2)x(x - R)
y = [4hx(R-x)] / R^2

 

[lyounamu]

equation of motion: y = - kx(x - R) ------(1)

sub (R/2 , h) to find k:
h = - kR/2 (R/2 - R)
4h/R^2 = k

sub k into (1):
y = - (4h/R^2)x(x - R)
y = [4hx(R-x)] / R^2

How did you get that equation for motion?

 

[lyounamu]

its just a general equation of a concave down parabola with roots at x = 0 (where the particle is fired) and x = R (where the particle lands).

remember that any parabola can be expressed as y = - a^2(x - A)(x - B)
(with my equation i made a^2 equal k)

Oh...

That was a brilliant thinking there. I never thought of that!!!