Question

CU-Urutan
A stone is projected from point A with speed u making an angle 60° with horizontal
as shown. The fixed inclined surface makes an angle 30° with horizontal. The stone
lands at B after time t. Then the distance AB is equal to
(A) ut
(B) v3ut
(C) But
(D) 2ut
.
.. ich nath
ubh​

Answer 1

Answer: (A)  ut/√3

Explanation:

1) We have,  a stone projected in a inclined plane as shown in figure.

t = time taken by stone to move from A to B.

We will convert this problem to normal ground projection with X-ray axis along the inclined plane and y-axis perpendicular to the inclined plane.

2) After manipulating with Angles as shown in figure ,we observered that

$a_x = -g sin(30\degree)\\ \\a_y = g cos(30\degree)\\ \\u_x =ucos(30\degree) \\ \\u_y = u sin(30\degree)$

3) We know,  

Time of flight is same 't'.

$t =\frac{2u_y}{|a_y|} \\ \\=\frac{2*usin(30\degree)}{gcos(30\degree)} \\ \\=\frac{2u}{g\sqrt{3}}$

4) We know,

Using newtons Equation of motion,  

$S_x = u_xt+1/2a_xt^2\\ \\=u cos(30\degree)-0.5*g sin(30\degree)*t^2\\ \\=\frac{u\sqrt{3}}{2}-0.5*g*0.5*t^2\\ \\=\frac{u\sqrt{3}}{2}-\frac{2u*0.5*0.5*t}{\sqrt{3}}\\ \\=\frac{u\sqrt{3}}{2}-\frac{u}{2\sqrt{3}}\\ \\=\frac{ut}{\sqrt{3}}$

Hence,  distance from A to B

= Sx = ut /√3 units.

Answer 2

Answer:

answer is option a only from which book is this