A good question

x = u²cosθ/g

z= u²(2sinθ - 1)/2g

is it rite?

no yaar this is not the answer!

there should be mass term too included!
this is not the answer!

hey then the mass of bullet shud be given !! if we have to apply conservation of linear p, they shud give the mass of the fired bullet !!

if we are simply to apply kinematics, then i think mass will not be involved newhr...

oh thats a blunder! i am sorry the mass of the bullet is m

body shoddy bana raha hai kya???????[3][3]impress everyone in IIT[9][9]

hhehehheheh [9]

wats gym???........

ohho prajith,,,,..u dont know what a gym is??????????[3][3]

no no eureka nothing like this! its all a part of daily routine! so i have to go for two hours, one hr in morning and one in evening! well the most important of all

KOI MUJHE ISKA SOLUTION DE DO!!!!!!!!!!!!!!

for the students who do not understand hindi,

RELP I NEED THE SOLUTION!!!!!!!!!!

well let the velocity of the combined system of machine gun and plank be v1 along negative x axis and v2 along y axis after the bullet is unleashed

then as the bullet was launched at a speed u w.r.t the gun the velocity of the bullet w.r.t the gun along y axis must be zero
(AS the gun is atr an uniform angle theta in x z plane
)

let the actual velocity of the gun be v3 with an angle of β to the x,z plane with x

Further as the bullet is launched perpendicular to the motion of the plank the relative velolcity along y must be zero

Mv=(M-m)v2+mvx

but vx=v2

thus we get v2=v and velocity of the bullet along y= v

further consercing momentum in the x-z plane

(M-m)v1=mv3cosβ

as the velocity of the bullet w.r.t plank is u ,its horizontal and vertical components must be ucos and usin theta

thuis

v3cosβ+v1=ucosθ
v3sinβ=usinθ

solving we get v3 cosβ and v3sinβ as the following

ucosθ-(mucosθ/M)
and usinθ

hence y coordinate=vt=vu/g
z coordinate= vsinβt-1/2gt²=u²/g(sinθ-1/2)
x coordinate=v3cosβt=u²cosθ/g(1-(m/M))

rohan is got the correct answer!