A cricketer can throw a ball to a maximum horizontal distance of 100m. To what height above the ground can the cricketer throw the same ball with...

Wednesday, June 5, 2013

A cricketer can throw a ball to a maximum horizontal distance of 100m. To what height above the ground can the cricketer throw the same ball with...

It is well known that a maximum distance for a given initial speed is reached if the angle is $\displaystyle{45}$ degrees above horizontal. Denote the initial speed as $\displaystyle{V}_{{0}}$ and the corresponding horizontal distance as $\displaystyle{D}_{{\max}}.$ Let's find $\displaystyle{V}_{{0}}.$

If thrown $\displaystyle{45}$ degrees above horizontal, the horizontal distance is $\displaystyle{D}{\left({t}\right)}=\frac{{{V}_{{0}}{t}}}{\sqrt{{{2}}}}$ and the height is $\displaystyle{H}{\left({t}\right)}=\frac{{{V}_{{0}}{t}}}{\sqrt{{{2}}}}-\frac{{{{g{{t}}}}^{{2}}}}{{2}}$ ( $\displaystyle\frac{{1}}{\sqrt{{{2}}}}$ is $\displaystyle{\cos{{\left({45}\right)}}}$ and $\displaystyle{\sin{{\left({45}\right)}}}$ ).

A ball falls at $\displaystyle{t}_{{1}}\gt{0}$ when $\displaystyle{H}{\left({t}_{{1}}\right)}={0},$ i.e. $\displaystyle{t}_{{1}}=\frac{{2}}{{g{\cdot}}}\frac{{V}_{{0}}}{\sqrt{{{2}}}},$ so $\displaystyle{D}{\left({t}_{{1}}\right)}=\frac{{{V}_{{0}}^{{2}}}}{{{g}}}={D}_{{\max}}.$ Thus $\displaystyle{V}_{{0}}=\sqrt{{{D}_{{1}}{g}}}.$

Great, now if a cricketer throws a ball vertically upward with the same initial speed, the maximum height of a ball would be $\displaystyle\frac{{{V}_{{0}}^{{2}}}}{{{2}{g}}}=\frac{{D}_{{\max}}}{{2}}.$ The simplest way to obtain this formula is to consider the energy conservation law, $\displaystyle{m}{g{{H}}}=\frac{{{m}{{V}_{{0}}^{{2}}}}}{{2}}.$

So the maximum height is twice less than the maximum horizontal distance, i.e. it is 50 m.

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Wednesday, June 5, 2013