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A bullet on penetrating $30 \mathrm{cm}$ into its target loses its velocity by $50 \%$. What additional distance will it penetrate into the target before it comes to rest?
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2875 Upvotes
Verified Answer
The correct answer is:
$10 \mathrm{cm}$
Given $S=30 \mathrm{cm}=30 \times 10^{-2} \mathrm{m}$ and loss in
velocity is $50 \%$ Equation of motion
$$
v^{2}=u^{2}+2 a s
$$
Here, $\frac{u^{2}}{4}=u^{2}+2 a \times 30 \times 10^{-2}$
$$
0=\frac{u^{2}}{4}+3 a \times x
$$
On solving Eqs. (i) and (ii), we get
$$
x=10 \mathrm{cm}
$$
velocity is $50 \%$ Equation of motion
$$
v^{2}=u^{2}+2 a s
$$
Here, $\frac{u^{2}}{4}=u^{2}+2 a \times 30 \times 10^{-2}$
$$
0=\frac{u^{2}}{4}+3 a \times x
$$
On solving Eqs. (i) and (ii), we get
$$
x=10 \mathrm{cm}
$$
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