shsshs said:
could sum1 please do the following?
1)
a particle is projected 30ms at the foot of an inclined plane which is
30[FONT=ＪＳ 明朝]∘ to the horizontal. For what angles of projection will the particle strike the incline plane
a) horizontally[/FONT]
b) perpendicularly
[FONT=ＪＳ 明朝]2) a vertical pole subtends an angle[FONT=ＪＳ 明朝]α at a point P on the ground. Two particles are projected simultaneously from P, making angles [FONT=ＪＳ 明朝]β and γ. The particle projected at [FONT=ＪＳ 明朝]β strikes the top of the pole and the instant the latter strikes the bottom. Prove that tan[FONT=ＪＳ 明朝]β- tanγ = tan[FONT=ＪＳ 明朝]α<O
</O
[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
Hi
shsshs,
Question 1a):
i) Both parts a) and b) involve statements about the
velocity(remember vectors:
directions as well) of the projectile at point of contact.
A natural question would then be "what is mathematically related to the direction/angle of velocity"?
A: The
gradient of the path of the projectile at point of contact.
ii) How does one go about finding the gradient at a particular point?
A: dy/dx.
Thus, we require the
equation of path of the projectile:
y = xTan@ - gx
2/(2(vCos@)
2) ; where @ is the angle of projection and v the initial
speed.
--->
dy/dx = Tan@ - gx/(vCos@)2 ____________________ (1)
Now, striking "horizontally" ---> dy/dx = 0 at point of contact.
---> 0 = Tan@ - gx/(vCos@)
2
--->
x = (v2Cos@Sin@)/g ____________________ (2)
To determine @, we need another expression for 'x' at point of contact. We get that from considering where the projectile strikes the incline:
a) The equation of the incline is: y = xTan(30
)
Substituting a) into the
equation of path to find point of contact:
xTan(30
) = xTan@ - gx
2/(2(vCos@)
2)
We can cancel out 'x' on both sides since x = 0 is a trivial solution which we are not interested in:
Tan(30
) = Tan@ - gx/(2(vCos@)
2)
--->
x = [2(vCos@)2(Tan@ - Tan(30))]/g ____________________ (3)
Solve (3) & (2) simultaneously:
[2(vCos@)
2(Tan@ - Tan(30
))]/g = (v
2Cos@Sin@)/g
---> 2Cos
2@(Tan@ - Tan(30
)) = Cos@Sin@
---> 2(Tan@ - Tan(30
)) = Tan@
---> Tan@ = 2Tan(30
) = 2/Sqrt(3)
i.e.
@ = ArcTan(2/Sqrt(3)) is the required angle of projection. (where "ArcTan" is the inverse tangent function)
Q1b):
We use the same argument as in Q1a). Thus, striking perpendicularly to the incline means dy/dx = -1/Tan(30
) =
-Sqrt(3) , since 'Tan(30
)' is the gradient of the incline.
Equating expression (1) with this yields:
-Sqrt(3) = Tan@ - gx/(vCos@)
2
--->
x = (vCos@)2(Tan@ + Sqrt(3))/g ____________________ (4)
Now, the second equation required to solve for @ is the same equation, (3), we used in Q1a) for the point of contact (since the setup does not change here).
Equating (3) and (4) above gives:
[2(vCos@)
2(Tan@ - Tan(30
))]/g = (vCos@)
2(Tan@ + Sqrt(3))/g
---> 2(Tan@ - Tan(30
)) = (Tan@ + Sqrt(3))
---> Tan@ = Sqrt(3) + 2Tan(30
) = Sqrt(3) + 2/Sqrt(3) = 5/Sqrt(3)
i.e.
@ = ArcTan(5/Sqrt(3)) is the required angle of projection.
Question 2:
Let P be the 'origin' in the frame of reference of the projections of the particles.
Let 'd' be the horizontal distance between P and the vertical pole along the ground.
Therefore, the vertical height 'h' of the pole is given by:
h = dTanα
At position x = d the difference between the heights of the particles is h. So using the
equations of path for both particles we get:
y
1 - y
2 = h = dTanα
---> dTanα = [dTanβ - gd
2/(2(v
1Cosβ)
2)] - [dTanγ - gd
2/(2(v
2Cosγ)
2)]
Cancel d from both sides as d doesn't = 0:
Tanα = [Tanβ - gd/(2(v
1Cosβ)
2)] - [Tanγ - gd/(2(v
2Cosγ)
2)]
i.e.
Tanα = Tanβ - Tanγ + (gd/2)[1/(v2Cosγ)2 - 1/(v1Cosβ)2] ____________________ (5)
Now, the second piece of information we know is that both particles strike the pole at their respective positions at the same instant t
0. So let's consider their respective horizontal displacements:
a) x
1 = v
1t
1Cosβ
b) x
2 = v
2t
2Cosγ
So at t
1 = t
2 = t
0, we know that x
1 = x
2 = d.
Equating a) and b) yields:
d = v
1t
0Cosβ = v
2t
0Cosγ
---> 1/(v
2Cosγ)
2 = 1/(v
1Cosβ)
2
i.e.
1/(v2Cosγ)2 - 1/(v1Cosβ)2 = 0 ____________________ (6)
Substitute (6) into (5) gives:
Tanα = Tanβ - Tanγ + (gd/2)[1/(v
2Cosγ)
2 - 1/(v
1Cosβ)
2]
= Tanβ - Tanγ + 0
----->
Tanα = Tanβ - Tanγ
as required.
Hope that helps.
