JEE QUESTIONS AND DOUBTS IN KINEMATICS-2

Choosing the conventional coordinate system, we have
v_{1x}(t) = v_1, v_{1y} = -gt which gives \tan \theta_1(t) = -\frac{gt}{v_1}

Similarly, \tan \theta_2(t) = \frac{gt}{v_2}

Since the vectors are perpendicular, we get \tan \theta_1(t)\tan \theta_2(t) = -1 \Rightarrow t = \frac{\sqrt{v_1v_2}}{g}

The vertical displacement is same for the particles and hence the distance between them is the same as the horizontal distance between them which is obviously (v_1+v_2)t = (v_1+v_2)\frac{\sqrt{v_1v_2}}{g}

yeah u r right Qwerty. thank you.

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 ?

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?

Yeah, i see. it's dimensionally incorrect. But this is what the ans is given