Notes: A soccer ball is kicked from the playing field at a 45° angle. If the ball is in the air for 3 s, what is the maximum height achieved?

Wednesday, 24 August 2016

A soccer ball is kicked from the playing field at a 45° angle. If the ball is in the air for 3 s, what is the maximum height achieved?

Hello!

Denote the angle as $\displaystyle\alpha,$ the initial speed as $\displaystyle{V}$ and the given time as $\displaystyle{T}.$

I suppose we ignore air resistance. Then the only force acting on the ball is the gravity force, it is directed downwards and gives the acceleration $\displaystyle{g{=}}{9.8}\frac{{m}}{{{s}}^{{2}}}$ to the ball.

Hello!

Denote the angle as $\displaystyle\alpha,$ the initial speed as $\displaystyle{V}$ and the given time as $\displaystyle{T}.$

The vertical component of the velocity uniformly decreases with time $\displaystyle{t}$ from $\displaystyle{V}{\sin{{\left(\alpha\right)}}}$ with the speed $\displaystyle{g},$ so it is equal to $\displaystyle{V}{\sin{{\left(\alpha\right)}}}-{g{{t}}}.$ The height itself is equal to $\displaystyle{H}{\left({t}\right)}={V}{\sin{{\left(\alpha\right)}}}{t}-\frac{{{{g{{t}}}}^{{2}}}}{{2}}.$ At the time $\displaystyle{T}$ the velocity is zero, i.e. $\displaystyle{V}{\sin{{\left(\alpha\right)}}}{T}=\frac{{{{g{{T}}}}^{{2}}}}{{2}},$ or $\displaystyle{V}{\sin{{\left(\alpha\right)}}}=\frac{{{g{{T}}}}}{{2}}.$

The maximum height is reached when the vertical speed becomes zero, i.e. when $\displaystyle{V}{\sin{{\left(\alpha\right)}}}={g{{t}}}.$ From the above we know that this time is $\displaystyle\frac{{T}}{{2}}.$

Finally, the maximum height is

$\displaystyle{H}{\left(\frac{{T}}{{2}}\right)}={V}{\sin{{\left(\alpha\right)}}}\frac{{T}}{{2}}-\frac{{{{g{{T}}}}^{{2}}}}{{8}}=\frac{{{{g{{T}}}}^{{2}}}}{{4}}-\frac{{{{g{{T}}}}^{{2}}}}{{8}}=\frac{{{{g{{T}}}}^{{2}}}}{{8}}.$

Numerically it is $\displaystyle\frac{{{9.8}\cdot{9}}}{{8}}\approx{11}{\left({m}\right)}.$ This is the answer. Note that it doesn't depend on $\displaystyle\alpha.$