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 â€‹Navya,

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) = 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

-UyT​ = 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.


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a projectile  is thrown  with velocity of 50 m/s  towards an inclined plane from ground such that it strikes the inclined plane perpendicularly the  angle of projection of the projectile 50 degree with the horizontal  and the inclined plane is inclined at an angle of  45 degree  to the horizontal . find the distance between the point of projection and the foot of  inclined plane .
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