## Statement

difficulty We have a small device that can launch missiles with a velocity of . Determine how far we must be to hit a target if:

• the height from which the launch occurs is 1.8 m
• the target is at a height of 1.5 m

## Solution

Data

• Initial velocity of the motion
• Initial height: y0 = H = 1.8 m
• Height of the target (final height) y = 1.5 m

Preliminary considerations

Resolution

The equation of position in parabolic motion is given by the equation:

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

The position vector at the end of the motion is given by:

$x\stackrel{\to }{i}+1.5·\stackrel{\to }{j}$

The unknown x, is just the asked value, the distance that we must be from the target for the projectile to hit it. Therefore, making the final position vector the same to the generic expression of the position vector, we can solve for the unknown that we need to find:

$\stackrel{\to }{r}\left(t\right)=\left({v}_{0x}·t\right)·\stackrel{\to }{i}+\left(H+{v}_{0y}·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}=x\stackrel{\to }{i}+1.5·\stackrel{\to }{j}$

Therefore:

Finally, knowing t, we solve for x

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