A projectile is thrown with velocity of 50 m/sec towards an inclined plane from ground such that it strikes the inclined plane perpendicularly the angle of projection of the projectile is 53^0 with the horizontal and the inclined plane is inclined at an angle of 45^0 to the horizontal.find the distance between the point of projection and the foot of inclined plane.
hi âNavya,
Please find below the solution to the asked query:
Taking the x axis along the inclined plane and y a xis perpendicular to it:
Resolving the given velocity alog our chosen co ordinate sysstem :
we have ,
Ux = 50Cos (53-45)o = 50Cos8o
Uy = 50Sin(53-45)o = 50Sin8o
similarlt the accelaration due to gravity can be resolved along the axis as
Ax = - gSin45o
Ay = -gCos45o
Applying equation of motion along the y axis:
As the body is projected from the inclined plane and finally falls on the inclined plane so,
Y = 0
hence,
using
Y = UyT + AyT2/2
putting Y= 0
we get
-UyTâ = AyT2
or , AyT = -Uy
T = -Uy/Ay .
T = 50Sin8o/gCos45o
T = (50 x 0.139)/(10 x 0.707)
T = 6.95 / 7.07 = 0.98 sec
Observing motion along positive x axis
we have :
Sx = UxT + AxT2/2
substitutuing the values :
or, Sx = 50Cos8o(0.98) - [gSin45o(0.98)2]/2
Sx = ( 49.5 x 0.98) - [7.07 x 0.96]/2
Sx = 48.51 - 3.39 = 45.12 m
This is distance between the point of projection and the foot of inclined plane.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Please find below the solution to the asked query:
Taking the x axis along the inclined plane and y a xis perpendicular to it:
Resolving the given velocity alog our chosen co ordinate sysstem :
we have ,
Ux = 50Cos (53-45)o = 50Cos8o
Uy = 50Sin(53-45)o = 50Sin8o
similarlt the accelaration due to gravity can be resolved along the axis as
Ax = - gSin45o
Ay = -gCos45o
Applying equation of motion along the y axis:
As the body is projected from the inclined plane and finally falls on the inclined plane so,
Y = 0
hence,
using
Y = UyT + AyT2/2
putting Y= 0
we get
-UyTâ = AyT2
or , AyT = -Uy
T = -Uy/Ay .
T = 50Sin8o/gCos45o
T = (50 x 0.139)/(10 x 0.707)
T = 6.95 / 7.07 = 0.98 sec
Observing motion along positive x axis
we have :
Sx = UxT + AxT2/2
substitutuing the values :
or, Sx = 50Cos8o(0.98) - [gSin45o(0.98)2]/2
Sx = ( 49.5 x 0.98) - [7.07 x 0.96]/2
Sx = 48.51 - 3.39 = 45.12 m
This is distance between the point of projection and the foot of inclined plane.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.