Projectiles On An Inclined Plane

[Woolyabyss]

Homework Statement

A particle is projected with initial speed u at an angle tan-inverse(1/2) to an inclined plane. The plane contains the point of projection and makes an angle tan-inverse(3/4) with the horizontal. Show that the angle at which the particle strikes the plane is tan-inverse(2)

Homework Equations

the line of greatest slope of the inclined plane is the taken as the horizontal(i axis) and the perpendicular of the line of greatest slope is taken as the j axis.

S=ut + 1/2(a)(t^2)

V = u + at

The Attempt at a Solution

I tried to find the angle which the object was projected to the plane. I called it A

tan(( tan-inverse(1/2) - tan-inverse(3/4) ) = 2/11 ... TanA =2/11

cosA = 11/(5√5) ...... sinA = 2/(5√5)

I tried to find the time when the object lands This is when the perpendicular distance from the plane is zero( Sy = 0)

Sy = u(2u / 5√5)t - (1/2)(4g/5)t^2

simplify and t = u/(g√5)

the angle when the object lands is given by TanB = A | (Vy)/(Vx) | (where the time is when the object lands.)

Vy
= u(2u/5√5) - g(4/5)(u/g√5)

= 2u/5√5 - 4u/5√5

= -2u/5√5


Vx = u(11/5√5) - g(3/5)(u/g√5)

simplify and Vx = 8u/5√5

when you divide Vy/Vx = 1/4

Im not sure what I did wrong but I think it might be something to do with "The plane contains the point of projection".
Any help would be appreciated.

 

[ehild]

A particle is projected with initial speed u at an angle tan-inverse(1/2) to an inclined plane. The plane contains the point of projection and makes an angle tan-inverse(3/4) with the horizontal. Show that the angle at which the particle strikes the plane is tan-inverse(2)

I tried to find the angle which the object was projected to the plane. I called it A

tan(( tan-inverse(1/2) - tan-inverse(3/4) ) = 2/11

You do not need to find that angle: it is given. Read the sentence in bold.


ehild

 

[Woolyabyss]

You do not need to find that angle: it is given. Read the sentence in bold.


ehild

Oh, For some reason I thought that it said it was the angle from the projection to the horizontal. I just did the question again using what you said and got the right answer. Thanks for the help.

 

[ehild]

You are welcome. Nice solution!

ehild

 

[Nickie]

Your approach to finding the angle at which the particle strikes the plane is correct, but there are a few errors in your calculations. First, when finding the time when the object lands, you should use the full expression for Sy, which is u(2u/5√5)t - (1/2)(4g/5)t^2. This will give you a quadratic equation, and you can use the quadratic formula to solve for t. Also, in your calculation of Vy, you have a mistake in your simplification. It should be 2u/5√5 - 4u/5√5 = -2u/5√5, not -4u/5√5. Finally, when finding the angle at which the object lands, you should use the full expression for Vx, which is u(11/5√5) - g(3/5)(u/g√5). This will give you a different value for Vx, and when you divide Vy/Vx, you should get tanB = 2, which is the desired result. Overall, your approach is correct, but you just need to be more careful with your calculations.