Question

A missile is moving 1810 m/s ata 20.0° angle. It needs to hit atarget 19,500 m away in a 32.0°direction in 9.20 s. What is themagnitude of the accelerationthat the engine must produce?

Answer 1

First let's calculate the vertical and horizontal components of the distance:

$\begin{gathered} dh=19500\cdot\cos (32\degree) \\ dh=16536.94 \\ \\ dy=19500\cdot\sin (32\degree) \\ dy=10333.43 \end{gathered}$

Then, since we have the time required, let's find the speed in each direction:

$\begin{gathered} vh=\frac{dh}{t} \\ vh=\frac{16536.94}{9.2} \\ vh=1797.49 \\ \\ vy=\frac{dy}{t} \\ vy=\frac{10333.43}{9.2} \\ vy=1123.2 \end{gathered}$

Now, let's decompose the given speed in its components:

$\begin{gathered} Vh=1810\cdot\cos (20\degree) \\ Vh=1700.84 \\ \\ Vy=1810\cdot\sin (20\degree) \\ Vy=619.06 \end{gathered}$