Statement

difficulty

A 1.95 m tall shot put athlete put the shot 25 meters away. Knowing that the path begins with an elevation of 40°, calculate:

1. The shot flight time
2. The shot initial velocity
3. The motion maximum height

Solution

Data

• Initial launching height y0 = H = 1.95 m
• Final distance on the horizontal axis x = 25 m
• Initial angle

Preliminary considerations

• On its trajectory, the shot will describe a parabolic motion, which is a composition of a uniform rectilinear motion in the x-axis, and a uniformly accelerated rectilinear motion in the y-axis
• The initial distance on the horizontal axis x0 = 0
• The x component of the initial velocity is given by ${v}_{0x}={v}_{0}·\mathrm{cos}\left(\alpha \right)$
• The y component of the initial velocity is given by ${v}_{0y}={v}_{0}·\mathrm{sin}\left(\alpha \right)$
• We consider the gravitational acceleration g = 9.8 m/s2

Resolution

The equation of position of the parabolic motion is given by the expression

$\stackrel{\to }{r}\left(t\right)=x\left(t\right)·\stackrel{\to }{i}+y\left(t\right)·\stackrel{\to }{j}=\left({x}_{0}+{v}_{0x}·t\right)·\stackrel{\to }{i}+\left(H+{v}_{0y}·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}$

1.

If the shot lands 25 m away it means that at that moment, its position vector is:

$\stackrel{\to }{r}\left(t\right)=25·\stackrel{\to }{i}+0·\stackrel{\to }{j}=\left(+{v}_{0}·\mathrm{cos}\left(\alpha \right)·t\right)·\stackrel{\to }{i}+\left(H+{v}_{0}·\mathrm{sin}\left(\alpha \right)·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}$

Substituting values and solving we have:

$\begin{array}{r}25={v}_{0}·\mathrm{cos}\left(0.69\right)·t\\ 0=1.95+{v}_{0}·\mathrm{sin}\left(0.69\right)·t-\frac{1}{2}·9.8·{t}^{2}\end{array}}\begin{array}{r}{v}_{0}=\frac{25}{\mathrm{cos}\left(0.69\right)·t}\\ 0=1.95+{v}_{0}·\mathrm{sin}\left(0.69\right)·t-\frac{1}{2}·9.8·{t}^{2}\end{array}}0=1.95+25·\mathrm{tan}\left(0.69\right)-\frac{1}{2}·9.8·{t}^{2}⇒\overline{)t=2.1s}$

2.

In relation to its components:

${\stackrel{\to }{v}}_{0}=\left({v}_{0}·\mathrm{cos}\left(0.69\right)\right)·\stackrel{\to }{i}+\left({v}_{0}·\mathrm{sin}\left(0.69\right)\right)·\stackrel{\to }{j}=\overline{)11.8·\stackrel{\to }{i}+·9.8\stackrel{\to }{j}}$

3.

The maximum height is reached when the velocity in the vertical y-axis is zero:

Substituting this value of t in the y component of the equation of position we get the maximum height value:

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas
Related sections
${v}_{x}={v}_{0x}={v}_{0}·\mathrm{cos}\left(\alpha \right)$
${v}_{y}={v}_{0y}-g\cdot t={v}_{0}·\mathrm{sin}\left(\alpha \right)-g\cdot t$
$x={v}_{x}\cdot t={v}_{0}·\mathrm{cos}\left(\alpha \right)·t$
$y=\mathrm{H}+{\mathrm{v}}_{0\mathrm{y}}·\mathrm{t}-\frac{1}{2}·g·{t}^{2}=\mathrm{H}+{\mathrm{v}}_{0}·\mathrm{sin}\left(\mathrm{\alpha }\right)·\mathrm{t}-\frac{1}{2}·g·{t}^{2}$