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3-2.Motion in Plane
normal
Two paper screens $A$ and $B$ are separated by a distance of $100\,m$. A bullet pierces $A$ and then $B$. The hole in $B$ is $10\,cm$ below the hole in $A$. If the bullet is travelling horizontally at the time of hitting $A$, then the velocity of the bullet at $A$ is $.......\,m / s$
A$100$
B$200$
C$600$
D$700$
Solution
(d)
$h=\frac{1}{2} g t^2$ (in vertical direction)
$\therefore \quad t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 0.1}{10}}=0.141\,s$
Now, in horizontal direction,
$v_x=\frac{S_x}{t}=\frac{100}{1.141}=700\,m / s$
$h=\frac{1}{2} g t^2$ (in vertical direction)
$\therefore \quad t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 0.1}{10}}=0.141\,s$
Now, in horizontal direction,
$v_x=\frac{S_x}{t}=\frac{100}{1.141}=700\,m / s$
Standard 11
Physics
