JEE QUESTIONS AND DOUBTS IN KINEMATICS-2

yeah u r right Qwerty. thank you.

1st one is very easy...

so (v1 i + gt j) (v2 i + gt j) = 0 ( as 2 vectors r perpendicular)

so t =√v1 v2 /g
distance = √v1 v2 /g (v1 + v2)

are ha , dont confuse x with horizontal coordinate, i have taken it as vertical coordinate only

i didnt get wich is the surface?

any way

while going up , net force = - (mg+cu²) ( j )

and while going down net force = (cu² - mg ) ( j )

now while going up ,

accⁿ a = mg+ cu²m = - dudt= - ududx

so

mudumg+cu² = -dx

i.e mclog(mg+cu²)= -x

where u changes from u_o to u , and x changes from 0 to x

(x measured from lowest point )

so, - cx = mlog(mg+cu²mg+cu_o²)

i m assuming that the surface is the one frm which the particle is thrown

while going down

accⁿ a = mg-cu²m = +dudt= + ududx

mudumg-cu² = dx

i.e
- mclog(mg-cu²) = x

where u changes from 0 to some u

and x changes from 0 to some x (measured from highest point )

so - mclog(mg-cu²mg) = x
so we got 2 eqns

- x = mclog(mg+cu²mg+cu_o²) ........(1)

- mclog(mg-cu²mg) = x ...........(2)

in eqn 1 , put u = 0 , and x =x_max

in eqn 2 , u is the u required , and put x = x_max

and equate x_max in the 2 eqns

so mgmg+cu_o² = mg-cu²mg

m²g² = m²g² +cmg(u_o²-u²) - c²u_o²u²

so mg(u_o²-u²)=cu_o²u², i.e mgu_o²=u²(cu_o²+mg)

u² = mgu_o²cu_o²+mg

wats the ans given ?

ya....

u sure???

THANX PROPHET SIR. PLEASE TRY THE NEXT ONE TOO

Some1 try these out, yaar, pleaaaase.

Some1 try this out...

air drag in the direction of W or opposite??

anyway, can u solve the second one??

i found it in the FIITJEE package( of one of my friends).

whats the source of ur quest?

for 1st the ans that u hav posted is dimensionally not correct.... plz check out