An inclined plane is making an angle   with horizontal  a projectile is projected  from  d bottom of d plane with a speed u at an angle   with horizontal den its range on d inclined plane is

[ans : R = 2u² Sin () cos / g Cos²()   ]

ive got d nr but i dont understand how v wud get cos²() 

den wats d time of flight & maximum range 2   plz

a point traversed 1/2 of d distance with a velocity v₀ d 1/2 of remaining part of d distance was  convered  with velocity v₁ & 2nd 1/2 of remaining part v₂ velocity d mean velocity of d point averaged over d whole time of motion is

ans : 2v₀(v₁+v₂) /(2v₀+v₁+v₂)

plz can u give me d explanation for d 1st quesn & 2nd too

look in 1 one we have to take the x dirrection along the incline and y direction prependicular to it .

U in x direction = u cos (a-b )    where b = beta and a = alpha

u in y direction = u sin(a-b)

also g will be resolved in two components

accleration in x direction = - g sin b

and in y direction = -gsin b

then in time t distance travelled in x direction

x= ucos (a-b) - 1/2 gsin b t^2   ------- 1

now we have to first find t

t can be obtained if y corrdinate = 0

0 =usin(a-b) t -1/2 gcos b t^2

t = 2usin(a-b) / gcos b

putting this in 1

we get

R = 2u^2 sin (a-b)cos a  /  gcos^2  b ... .

Akshay gave the derivation for time of flight and range,To find max. range,we proceed like this,Apply,2sinAcosB=sin(A+B)+sin(A-B),we get,

R=(u²/gcos²b)[sin(2a-b)-sinb] is max iff 2a-b=pi/2 and R_max=(u²/gcos²b)[1-sinb]=u²/g(1+sinb)

Here,b=beta and a=alpha.