Question

A projectile is thrown with velocity 50m/s towards an inclined plane from ground such that it strikes the inclined plane perpendicular. The angle of projection of projectile is 53° with the horizontal and inclined plane is inclined at an angle of 45° to the horizontal. Find (a) time of flight (b) the distance between the point of projection and the foot of inclined plane.

Answer 1

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) = 50Cos8
Uy = 50Sin(53-45) = 50Sin8
similarlt the accelaration due to gravity can be resolved along the axis as
Ax = - gSin45
Ay = -gCos45
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 = U T + A T /2
putting Y= 0
we get
-U T​ = A T
or , A T = -U
T = -U /Ay .
T = 50Sin8 /gCos45
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 :
S = U T + A T /2
substitutuing the values :
or, S = 50Cos8 (0.98) - [gSin45 (0.98) ]/2
S = ( 49.5 x 0.98) - [7.07 x 0.96]/2
S = 48.51 - 3.39 = 45.12 m
This is distance between the point of projection and the foot of inclined plane.


This is the Answers.