Question

A missile is moving 1350 m/s at a 25 degree angle. It needs to hit a target 23,500 m away in a 55 degree direction in 10.20 seconds. What is the direction of its final velocity?

Answer 1

`V = 1419.46\ \^x+3203.97\ \^y\ \ m/s`

Or

`|V|=3504.32\ m/s` with
`\alpha_x=66\°`

Step-by-step explanation:

We know that the initial speed is:

1350 m/s at a 25 degree angle

And the final position after 10.20 sec is:

23,500 m away in a 55 degree direction

In this problem we have:

Final position in x:

`23,500cos(55\°)=V_0x(t) + 0.5a_xt^2`

Where
`V_(0x)` is the initial velocity in x

So:

`23,500cos(55\°)=1350cos(25\°)(10.20) + 0.5a_x(10.20)^2`

We solve for
`a_x`

`23,500cos(55\°)-1350cos(25\°)(10.20) = 0.5a_x(10.20)^2`

`a_x=(23,500cos(55\°)-1350cos(25\°)(10.20))/(0.5(10.20^2))`

`a_x=19.21\ m/s^2`

X component of velocity

`Vx = 1350cos(25\°) + a_x(t)`

Where
`a_x` is the acceleration in the direction x

And
`a_x=19.21\ m/s^2`

Therefore:

`Vx = 1350cos(25\°) + 19.21(10.20)`

`Vx = 1419.46\ m/s`

Final position in y:

`23,500sin(55\°)=V_0y(t) + 0.5a_yt^2`

Where
`V_(0y)` is the initial velocity in y

So:

`23,500sin(55\°)=1350sin(25\°)(10.20) + 0.5a_y(10.20)^2`

We solve for
`a_y`

`23,500sin(55\°)-1350sin(25\°)(10.20) = 0.5a_y(10.20)^2`

`a_y=(23,500sin(55\°)-1350sin(25\°)(10.20))/(0.5(10.20^2))`

`a_y=258.18\ m/s^2`

Y component of velocity

`V_y = 1350sin(25\°) +a_y(t)`

Where
`a_y` is the acceleration in the direction y

And
`a_y=258.18\ m/s^2`

Therefore:

`Vy = 1350sin(25\°) + 258.18(10.20)`

`Vy = 3203.97\ m/s`

Note then that the final velocity V is a two component vector

`V = Vx+Vy`

`V = 1419.46\ \^x+3203.97\ \^y\ \ m/s`

The magnitude of the final speed is:

`|V| =√(V_x^2+V_y^2)\\\\|V|=√((1419.46)^2+(3203.97)^2)\\\\|V|=3504.32\ m/s`

And the angle that forms with the x axis is:

`\alpha=cos^(-1)((V_x)/(|V|))\\\\\\\alpha =cos^(-1)((1419.46)/(3504.32))\\\\\alpha=66\°`