TS EAMCET 20-July-2022 Shift 2 Solved Paper

Given situation is shown below.
As it is clear from above diagram, velocity perpendicular to incline =x‌cos‌30∘ or u⟂=u‌cos‌30∘ (upward).
=10×‌

√3

2

=5√3ms−1
And acceleration perpendicular to inclined plane is
a⟂‌=−g‌cos‌30∘‌ or ‌−gsin‌60∘
‌=−g×‌

√3

2

‌ (downwards) ‌
‌=−5√3m∕ s2
Let projected particle again lands on incline in time t, then
So, times of flight t is given by
‌s=u⟂t+‌

1

2

a⟂t2
‌⇒‌‌0=5√3t−‌

5√3

2

×t2
‌⇒‌‌t=0 or t=2 s
In two seconds particle lands on the incline. Initial velocity, x||=x‌cos‌60∘=‌

x

2

m/ s=5m/ s acceleration down inline,
a||=g‌cos‌60∘=‌

g

2

m/ s2=5m/ s2
Now, using s=ut+‌

1

2

at2
We get, ‌‌s=5×2+‌

1

2

×5×22=20m
So, particle covers 20m down the plane.