Question

Find the minimum initial speed of a red wine cork that travels a maximum horizontal distance of 12 m.

Answer 1

Let the cork's initial velocity be \(v_i\), and the angle at which it was fired above the horizontal be \(\theta\). We will assume that it lands at the same height it gets fired from. Since gravitational acceleration is given by magnitude \(g\) in the downward direction (and is the only acceleration acting on the cork if we neglect air friction), we can express the following about the location of the cork \((x,y)\) relative to its initial position.\[x = v_it \cos \theta\]\[y = v_it \sin \theta - \frac{1}{2}gt^2\]At the end of the trajectory, \(y=0\). Let us substitute this and find out at what time(s) this will occur.\[0=v_it \sin \theta - \frac{1}{2}gt^2\]\[0 = t\left(v_i \sin \theta - \frac{1}{2}gt\right)\]\[t=\left\{0,\frac{2v_i\sin\theta}{g}\right\}\]Let us plug the nontrivial (second) time into the equation for \(x\) so we can derive a expression for how far the cork will have gone \(\Delta x\) as a function of \(\theta\).\[\Delta x=\frac{2v_i^2 \cos \theta \sin \theta}{g}\]The keen physicist will recognize \(\sin 2\theta = 2\sin \theta \cos \theta\). Substituting this gives the following.\[\Delta x=\frac{v_i^2\sin 2 \theta}{g}\]It is clear that the maximum distance traveled for a given \(g\) and \(v_i\) is when \(\theta = \frac{\pi}{4}=45^\circ\). This causes \(\sin 2\theta = 1\).\[\Delta x=\frac{v_i^2}{g}\]Solving for \(v_i\) gives us our symbolic answer. Plug in given values to find \(v_i\).\[\boxed{\displaystyle v_i = \sqrt{g\Delta x}}\]

Answer 2

Plug in gravitational acceleration \(g\) and horizontal displacement \(\Delta x\). Use a calculator.