S.k.patel (1 Viewer)

[pepe]

I Need Help With Question 1 From Page 249- Ex.8a From This Textbook
(SECOND EDITION) -- The question is a bit too long.

 

[who_loves_maths]

Hi pepe,

could you plz specify which edition of the book you are referring to? there are different editions, and the one i have in front of me on pg 249 is Ex. 8B, not Ex. 8A.

or perhaps, if it's not too inconvenient, you could type up the question. (provided it's not TOO long )

 

[pepe]

pepe

sorry, second edition

 

[pepe]

pep

sorry again, you are right! is Ex. 8B - question 1

 

[who_loves_maths]

^ lol, okay, npz.

Question 1:

Part (i)

in order to strike the inclined plane horizontally, it means that the stone, at the point of contact with the plane, has ONLY horizontal velocity, and NO components of vertical velocity.
this only happens when the stone reaches the apex (or highest point) of its flight - the vertex of the parabola:

hence the max height is where the point of contact occurs: H = (v^2*Sin(a)^2)/2g
this height can also be calculated from the plane which is inclined at 30 degrees: H = Tan(30)*x ; where 'x' is the horizontal displacement of the stone at that point.
but at that point, the stone is at its peak ---> so x = R/2 , where 'R' is the range. ie. x = (v^2*Sin(2a))/2g

so equating both expressions:
(v^2*Sin(a)^2)/2g = (Tan(30)*v^2*Sin(2a))/2g ---> Sin(a)^2 = Sin(2a)/sqrt3 ; where a > 30
---> 0 = Sin(a)[Sin(a) - 2Cos(a)/sqrt3] ; Sin(a) can't = 0, as a > 30 ;
hence, Sin(a) = 2Cos(a)/sqrt3 ---> Tan(a) = 2/sqrt3 ---> ie. a = ArcTan(2/sqrt3)


Part (ii)

"at right angles" means that the velocity of the stone as it hits the inclined plane is perpendicular to the length of the inclined plane.

this means that the point on the parabolic path of the stone's flight at the point of contact has gradient -1/tan(30) {ie. since it's perpendicular) = -sqrt3

now, the equation of path is:
y = xtan(a) - gx^2/(2v^2*cos(a)^2) ; therefore, dy/dx = tan(a) - gx/(v^2*cos(a)^2) ; but dy/dx gives the gradient:
ie. -sqrt3 = tan(a) - gx/(v^2*cos(a)^2) ---> x = (tan(a) + sqrt3)(v^2*cos(a)^2/g)

the point of contact is also the intersection of the parabola and the line, so equate both:
tan(30)*x = y = xtan(a) - gx^2/(2v^2*cos(a)^2) ---> x = 2(tan(a) -(1/sqrt3))(v^2*cos(a)^2/g)

now, equate both expressions for 'x':
(tan(a) + sqrt3)(v^2*cos(a)^2/g) = x= 2(tan(a) -(1/sqrt3))(v^2*cos(a)^2/g)
---> tan(a) = sqrt3 + 2/sqrt3 = 5/sqrt3 ---> ie. a = ArcTan(5/sqrt3)


hope this helps

{note: there might be easier ways of doing these problems.}

 

[pepe]

Ex 8B-Q1

Thank you very much !!!!!