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A bullet of mass $10 \mathrm{~g}$ is fired horizontally with a velocity $1000 \mathrm{~ms}^{-1}$ from a rifle situated at a height $50 \mathrm{~m}$ above the ground. If the bullet reaches the ground with a velocity $500 \mathrm{~ms}^{-1}$, the work done against air resistance in the trajectory of the bullet is : $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$3750 \mathrm{~J}$
From equation of motion,
$v^2=u^2-2 a s$
$(500)^2=(1000)^2-2 \times a \times s$
$s=\frac{(1000)^2-(500)^2}{2 a}=\frac{375000}{a}$
$\therefore$ Work done against air resistance $=F S$
$=m a \times s$
$=\frac{10}{1000} a \times \frac{375000}{a}$
$=3750 \mathrm{~J}$
$v^2=u^2-2 a s$
$(500)^2=(1000)^2-2 \times a \times s$
$s=\frac{(1000)^2-(500)^2}{2 a}=\frac{375000}{a}$
$\therefore$ Work done against air resistance $=F S$
$=m a \times s$
$=\frac{10}{1000} a \times \frac{375000}{a}$
$=3750 \mathrm{~J}$
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