Answer to Question #137557 in Mechanics | Relativity for Fatima Binte Tariq

Question

The maximum speed with which you can throw a stone is about 25 m/s (a professional baseball pitcher can do much better than this).Can you hit a window 50 m away and 13 m up from the point where the stone leaves your hand? What is the maximum height of a window you can hit at this direction?

Answer 1

$x=vtcosα \Rightarrow t=\frac{x}{vcosα}$,

$y=vtsinα-\frac{gt^2}{2}=v\frac{x}{vcosα}sinα-\frac{gx^2}{2v^2cos^2α}=xtanα-\frac{g}{2v^2cos^2α}x^2$ - the equation of the parabola.

$y_{\alpha}^{'}=\frac{x}{cos^2α}-\frac{gx^2}{2v^2}\cdot\frac{2sinα}{cos^3α}=\frac{x}{cos^2α}\cdot (1-\frac{gxtanα}{v^2}),$

$y_{\alpha}^{'}=0 \Rightarrow 1-\frac{gxtanα}{v^2}=0 \Rightarrow tanα=\frac{v^2}{gx},$

$v=25 (\frac{m}{s}), x=50 (m), tanα=\frac{25^2}{10\cdot50}=\frac{5}{4} \Rightarrow cos^2α=\frac{1}{1+tan^2α}=\frac{1}{1+\frac{25}{16}}=\frac{16}{41}$.

$y=50\cdot\frac{5}{4}-\frac{10\cdot50^2}{2\cdot25^2\cdot\frac{16}{41}}=\frac{50\cdot5}{4}-\frac{5\cdot4\cdot41}{16}=\frac{250}{4}-\frac{205}{4}=\frac{45}{4}=11.25(m)$ - the maximum height of a window at the distance 50 m.

It means that we can't hit the window on a height of 13 m.