tan(15°) =___________

Answer

A, B

Explanation

Let the velocity of `A=mu_1`
Let the fiN/Al velocity when reachign at B become collision `=v_1`
`:.(1/2)m_1^2-(1/2)mv_1^2=mgh`
`rarr v_1^2-v_1^2=2gh`
`rarr v_1=sqrt(2gh+u_1^2)`……..i
When the block B reached at the upper man's hed, the velocity of B is just zero.
`:. (1/2)x2mxx(0)^2-(1/2)xx2mxxv^2=mgh`
rarr `v=sqrt(2gh)`
`:. Before collison velocity of u_A=v_1` ltbr. `u_B=0`
After collision velocity fo v_A=V(say)`
`v_B=sqrt(2gh)`
since it is an elastic collision the momentum and K be conserved
`:.mxxv_1+2mxx0=mxxv+2m+sqrt(2gh)`
`rarr v_1-v=2sqrt(2gh)`
Also `(1/2)xxmxxv_1^2+(1/2)xx2m(sqrt(2gh)^2)`
`rarr v_1^2-v^2=2xxsqrt(2gh)xx2msqrt(2gh)`
Dividing i and ii
`((v_1+v)(v_1-v))/(v_1+v)=(2xxsqrt(2gh)xsqrt(2gh))/(2xxsqrt(2gh))`
`rarr v_1+v=sqrt(2gh)`
dividing i and iii
`2v_1=3sqrt(2gh)`
`rarr v_1=(3/2)sqrt(2gh)`
`But v_1=sqrt(2gh+u^2)`
`=(3/2)sqrt(2gh)`
rarr 2gh+u^2=(9/4)(2gh)`
`u=sqrt(2.5gh)`
So the block will travel with a velocity greater than `sqrt(2.5gh)` i.e., man must awake.