First slide

Projection Under uniform Acceleration

Question

Two thin wood screens A and B are separated by 200 m. A bullet travelling horizontally at a speed of 600 ms^-1 hits the screen A, penetrates through it and finally emerges out from B making holes in A and B. If the resistance of air and wood are negligible, the difference of heights of the holes in A and B is

Easy

A

2 m
B

$\frac{49}{90} m$
C

$\frac{7}{\sqrt{90}} m$
D

zero

Solution

Here R = 200
U = 600
$\begin{array}{l} R = U \sqrt{\frac{2 h}{g}} \\ 200 = 600 \sqrt{\frac{2 \times h}{9 - 8}} \\ \frac{1}{9} = \frac{2 h}{9 .8}, h = \frac{4 .9}{9} = \frac{49}{90} m . \end{array}$

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