(i) show that tanx=2tanB. (ii) if the particle strikes the plane when moving in the direction perpendicular to the plane, show that 2tan(a-B)tanB=1

Question

a particle is projected at an angle alpha (a) to the horrizontal from a point on a plane inclined to angle beta (B) to the horrizontal. The path of the particle lies in a vertical plane through a line of greatest slope of the plane. The particle strikes the plane when moving horrizontally.

Answer 1

tanx in the question part (i) should be tana.

Equations of motion for the speed of projection, u, in terms of time, t, and projection angle, a.

Vertical: y=utsin(a)-gt^2/2=t(usin(a)-gt/2); horizontal: x=utcos(a).

y/x=(usin(a)-gt/2)/ucos(a)=tan(a)-gt/2ucos(a).

y'=dy/dt=usin(a)-gt; x'=dx/dt=ucos(a).

y'/x'=(usin(a)-gt)/ucos(a)=tan(a)-gt/ucos(a).

(i)

When y'=0, parabolic trajectory is horizontal and gt=usin(a), so t=usin(a)/g at the highest point (vertex).

At t=usin(a)/g, tan(B)=y/x=(usin(a)-usin(a)/2)/ucos(a)=tan(a)/2.

Therefore tan(a)=2tan(B).

(ii)

Slope of normal to the plane has to be -1/tan(B)=-cot(B).

y'/x'=-cot(B)=tan(a)-gt/ucos(a).

gt=(sin(a)/cos(a)+cos(B)/sin(B))ucos(a)=u(sin(a)sin(B)+cos(a)cos(B))/sin(B)=ucos(a-B)/sin(B).

y/x=tan(B) at this time: tan(B)=tan(a)-cos(a-B)/2cos(a)sin(B).

cos(a-B)/2cos(a)sin(B)=tan(a)-tan(B)=(sin(a)cos(B)-sin(B)cos(a))/cos(a)cos(B)=sin(a-B)/cos(a)cos(B).

So, cos(a-B)/2sin(B)=sin(a-B)/cos(B); cos(B)cos(a-B)=2sin(B)sin(a-B) and 2tan(B)tan(a-B)=1.

Comments

Thank you very much. But I need the rest please

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Still working on it and shortening the working out. Ran out of time yesterday. I can make a few shortcuts to reduce the text size which is becoming unmanageable and difficult to edit. Hope to have a concise answer soon, Oxygen1244. Thanks for your patience.

Complete solution now ready for viewing.

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Thank u very much. In fact, you are a mathematician