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=(dh)/(t) \\ vh=(16536.94)/(9.2) \\ vh=1797.49 \\ \\ vy=(dy)/(t) \\ vy=(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}`