A baseball is thrown vertically upward with an initial velocity of 14 mps. Find the: a) velocity with which it strikes the ground b) maximum...

Thursday, September 18, 2014

A baseball is thrown vertically upward with an initial velocity of 14 mps. Find the: a) velocity with which it strikes the ground b) maximum...

Hello!

The movement of a body thrown vertically upwards is uniformly accelerated (or decelerated, if you wish). The speed $\displaystyle{V}$ of a body is equal to $\displaystyle{V}{\left({t}\right)}={V}_{{0}}-\gt,$ where $\displaystyle{t}$ is a time, $\displaystyle{V}_{{0}}$ is an initial upward speed and $\displaystyle{g{\approx}}{9.8}\frac{{m}}{{{s}}^{{2}}}$ is the gravity acceleration. The minus sign before $\displaystyle{g{}}$ reflects the fact that gravity acceleration is directed downwards.

The height of a body is $\displaystyle{H}{\left({t}\right)}={H}_{{0}}+{V}_{{0}}{t}-\frac{{{{g{{t}}}}^{{2}}}}{{2}},$ where $\displaystyle{H}_{{0}}$ is the initial height (zero in our problem). A baseball strikes the ground at $\displaystyle{t}_{{1}}\gt{0}$ such that $\displaystyle{H}{\left({t}_{{1}}\right)}={0}.$ It is obvious that $\displaystyle{t}_{{1}}=\frac{{{2}{V}_{{0}}}}{{g{.}}}$ The velocity at this moment is $\displaystyle{V}{\left({t}_{{1}}\right)}={V}_{{0}}-{g{\cdot}}\frac{{{2}{V}_{{0}}}}{{{g}}}=-{V}_{{0}},$ i.e. the same but downwards.

The maximum height is reached at the parabola vertex, and the time is $\displaystyle{t}_{{2}}=-\frac{{b}}{{{2}{a}}}=\frac{{V}_{{0}}}{{{g}}}\approx{1.43}{\left({s}\right)}$ (the half of $\displaystyle{t}_{{1}},$ actually). The height at that time is $\displaystyle\frac{{{\left({V}_{{0}}\right)}}^{{2}}}{{{2}{g}}}\approx{10}{\left({m}\right)}.$