IIT JEE , AIEEE Forum - Physics , Mechanics

gopi_mith

Sir,

can anyone explain me the concepts and formulae and analysis of the projectile motion in an inclined plane

vpunithreddy

there is no difference b/n projectile in the horizontal plane and projectile in the inclined plane.the notation are taken with respect to inclined plane angle and then substitute them in the equation of the horizontal projectile

edison

PROJECTILE FIRED FROM AN INCLINED PLANE

Consider an inclined palne which makes an

with the horizontal. Let a projectile be projected with a velocity 'v' making an angle

with the horizontal.

Let us choose X axis along the inclined plane and Y-axis perpendicular to it.

gcos

and gsin

are the two rectangular components of g.

Now,

v_x= v cos(

-

) and v_y = v sin(

-

).

Let T = time of flight of the projectile

The displacement of the projectile perpendicular to the inclined plane is clearly

Zero.

Using S = ut + (1/2) a t², for motion along y-axis, we get

0 = v sin(

-

) T -(1/2) g cos

T²

or T = 2 v sin(

-

)/g cos

Range on the inclined plane is

R = 2v² sin(

-

) cos

/g cos²

R_max= v²/g (1+sin

)

H_max= v²sin²(

-

)/2g cos

If

is changed to -

, then we get the formula of time of flight and range for

'down the inclined plane'. In this case,

T = 2v sin(

+

)/g cos

R = 2v² sin(

+

) cos

/g cos²

R_max= v²/g (1-sin

)

cvramana

Consider the angle of inclination to be a and the angle of projection to be b with respect to the ground. Also let the velocity of projection be V. Consider X axis along the incline and Y axis perpendicular to the incline plane.

Therefore V_x = V cos (b - a) and V_y = V Sin (b - a)

g_x = g Sin (a) and g_y = g cos (a)

Y = V_y t - 0.5 g_y t ²

X = V_x t - 0.5 g_x t ²

Solving the equations we get the required results.

Time of flight can be obtained by putting Y = 0;

After obtaining T the time of flight, substitute T in X and obtain the range. Check for its maximum value.

neeraj

Let us consider an inclined plane which makes an angle b with the horizontal. Let a particle projected at angle a with horizontal with a velocity u. This particle strikes an inclined plane at a point A. Our aim is to find out time of flight and range OA

The initial velocity can be resolved into two components

(i) u cos (a - b) along the plane

(ii) u sin (a - b) perpendicular to the plane

Similarly g can be resolve in two components

(i) g sinb parallel to the plane (retardation)

(ii) g cosb perpendicular to the plane

If t be the time taken by the particle to go from O to A then in this time the distance described perpendicular to OA is zero

S = u t + ½ g t²

0 = u sin (a - b) - ½ g cos b. t²

t = [ 2 u sin (a - b)] / [g cos b ] -----.(1)

eq.1 represents time of flight

During time t horizontal velocity u cos a along OB remains const.

Distance traveled OB is given by

OB = ( u cos a) . t

= (u cos a) [ 2 u sin (a - b)] / g cos b

= [ 2 u² sin (a - b) . cos a ] / g cos b ---.. (2)

OA = OB / cos b = = [ 2 u² sin (a - b) . cos b ] / g cos² b

magiclko

well i think it will be better to get the axes rotated by an the angle of inclination.... as tym is frame independent... we can resolve the acceleration into "gsin

" along the incline n "gcos

" perpendicualr to it..... n then slove using the basic equations of motion.... n then the parameters which r to be derived can be evaluated back in the original frame.... x-y axis.....

how's it???? it will definetly save tym ....

neeraj

Hi,

There are many methods for solving above problem. You can choose whatever is convenient to you

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