TS EAMCET 18-July-2022 Shift 1 Solved Paper

Let u be initial velocity of bullet,

Final velocity of bullet after travelling a distance x is equal to (u−‌

u

3

)=‌

2u

3

We know, v2=u2+2as
⇒‌‌(‌

2u

3

)2=u2+2ax
⇒‌‌‌

4u2

9

−u2=2ax
⇒‌‌x=‌

−5

18a

u2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Again after embedding further distance of x′ , it comes to rest

v′2=v2+2ax′
(Now, v= initial velocity for x′ distance)
0=‌

4u2

9

+2×a×x′
⇒‌‌x′=‌

−4u2

18a

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
‌

x′

x

=‌

‌ −4u2 18a

−5u2 18a

=‌

4

5

[from Eqs. (i) and (ii)]