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Two-paper-screens-A-and-B-are-separated-by-100-m-distance-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-A-calculate-the-veloc




Question Number 15599 by Tinkutara last updated on 12/Jun/17
Two paper screens A and B are  separated by 100 m distance. 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 A, calculate  the velocity of the bullet at A.  (Neglecting the types of frictional  forces)
$$\mathrm{Two}\:\mathrm{paper}\:\mathrm{screens}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{100}\:\mathrm{m}\:\mathrm{distance}.\:\mathrm{A}\:\mathrm{bullet} \\ $$$$\mathrm{pierces}\:{A}\:\mathrm{and}\:\mathrm{then}\:{B}.\:\mathrm{The}\:\mathrm{hole}\:\mathrm{in}\:{B}\:\mathrm{is} \\ $$$$\mathrm{10}\:\mathrm{cm}\:\mathrm{below}\:\mathrm{the}\:\mathrm{hole}\:\mathrm{in}\:{A}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{bullet} \\ $$$$\mathrm{is}\:\mathrm{travelling}\:\mathrm{horizontally}\:\mathrm{at}\:{A},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bullet}\:\mathrm{at}\:{A}. \\ $$$$\left(\mathrm{Neglecting}\:\mathrm{the}\:\mathrm{types}\:\mathrm{of}\:\mathrm{frictional}\right. \\ $$$$\left.\mathrm{forces}\right) \\ $$
Answered by mrW1 last updated on 12/Jun/17
y=−(1/2)gt^2   x=ut  ⇒y=−(1/2)g((x/u))^2 =−(g/(2u^2 ))x^2   y=−10 cm at x=100 m  −0.1=−(g/(2u^2 ))×100^2   u=100(√(5g))=100(√(5×9.81))=700.4 m/s
$$\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{ut} \\ $$$$\Rightarrow\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\frac{\mathrm{x}}{\mathrm{u}}\right)^{\mathrm{2}} =−\frac{\mathrm{g}}{\mathrm{2u}^{\mathrm{2}} }\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{y}=−\mathrm{10}\:\mathrm{cm}\:\mathrm{at}\:\mathrm{x}=\mathrm{100}\:\mathrm{m} \\ $$$$−\mathrm{0}.\mathrm{1}=−\frac{\mathrm{g}}{\mathrm{2u}^{\mathrm{2}} }×\mathrm{100}^{\mathrm{2}} \\ $$$$\mathrm{u}=\mathrm{100}\sqrt{\mathrm{5g}}=\mathrm{100}\sqrt{\mathrm{5}×\mathrm{9}.\mathrm{81}}=\mathrm{700}.\mathrm{4}\:\mathrm{m}/\mathrm{s} \\ $$
Commented by Tinkutara last updated on 12/Jun/17
But answer is 700 m/s.
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{700}\:\mathrm{m}/\mathrm{s}. \\ $$
Commented by mrW1 last updated on 12/Jun/17
I had a mistake. now it′s fixed.
$$\mathrm{I}\:\mathrm{had}\:\mathrm{a}\:\mathrm{mistake}.\:\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{fixed}. \\ $$

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