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A bullet fired from a rifle has an initial velocity of \(v=700\text{ ms}^-1\). Drag force from the air on the bullet causes an acceleration of \(-3v^2\). How long will this bullet take to reach a target \(1500\text{ m}\) away horizontally? Clarification: There are no other forces acting on the bullet.

Note by Jerry Han Jia Tao
1 year, 5 months ago

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\[\begin{align}\frac{\mathrm{d}v}{\mathrm{d}t} &= -3 v^2 \\\implies \frac{1}{v}- \frac{1}{700}&=3t \\\implies x &= \frac{1}{3} \ln (1+2100t)\\ \text{Given: } x&=1500\, m \\\implies t &\approx 10^{1951} \ s\end{align}\]

Deeparaj Bhat - 1 year, 5 months ago

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Do we neglect gravity?

Deeparaj Bhat - 1 year, 5 months ago

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Yes.

Jerry Han Jia Tao - 1 year, 5 months ago

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I do not understand what v is in the acceleration. Just use the second equation of motion and solve.

Brilliant Member - 1 year, 5 months ago

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v is the speed of the bullet in that moment

Matteo Monzali - 1 year, 5 months ago

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