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.

## Comments

Sort by:

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

Log in to reply

Do we neglect gravity? – Deeparaj Bhat · 1 year, 2 months ago

Log in to reply

– Jerry Han Jia Tao · 1 year, 2 months ago

Yes.Log in to reply

I do not understand what v is in the acceleration. Just use the second equation of motion and solve. – Svatejas Shivakumar · 1 year, 2 months ago

Log in to reply

– Matteo Monzali · 1 year, 2 months ago

v is the speed of the bullet in that momentLog in to reply