Launching distance in parabolic motion

Data

• Initial velocity of the motion $\overrightarrow{v_0}=3·\overrightarrow{i}+4·\overrightarrow{j} m/s$ 
• Initial height: y₀ = H = 1.8 m
• Height of the target (final height) y = 1.5 m

Preliminary considerations

• It is a parabolic motion. The parabolic motion is a composition of two motions:
  • uniform rectilinear motion in the x-axis
  • uniformly accelerated rectilinear motion in the y-axis

Resolution

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

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

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

$$x\overrightarrow{i}+1.5·\overrightarrow{j}$$

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:

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

Therefore:

$$\begin{array}{l}\begin{array}{l}H+v_{0y}·t-\frac{1}{2}·g·t^2=1.5\Rightarrow 1.8+4·t-\frac{1}{2}·9.8·t^2\\ 0.3+4·t-\frac{1}{2}·9.8·t^2=0\Rightarrow -4.9·t^2+4·t+0.3=0\end{array}\\ t_1=0.88 s;\xcancel{ t_2=-}\end{array}$$ 

Finally, knowing t, we solve for x

$$x=v_{0x}·t=3·0.88=\boxed{2.65 m}$$