Answer:
[tex]V = 1419.46\ \^x+3203.97\ \^y\ \ m/s[/tex]
Or
[tex]|V|=3504.32\ m/s[/tex] with [tex]\alpha_x=66\°[/tex]
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:
[tex]23,500cos(55\°)=V_0x(t) + 0.5a_xt^2[/tex]
Where [tex]V_{0x}[/tex] is the initial velocity in x
So:
[tex]23,500cos(55\°)=1350cos(25\°)(10.20) + 0.5a_x(10.20)^2[/tex]
We solve for [tex]a_x[/tex]
[tex]23,500cos(55\°)-1350cos(25\°)(10.20) = 0.5a_x(10.20)^2[/tex]
[tex]a_x=\frac{23,500cos(55\°)-1350cos(25\°)(10.20)}{0.5(10.20^2)}[/tex]
[tex]a_x=19.21\ m/s^2[/tex]
X component of velocity
[tex]Vx = 1350cos(25\°) + a_x(t)[/tex]
Where [tex]a_x[/tex] is the acceleration in the direction x
And [tex]a_x=19.21\ m/s^2[/tex]
Therefore:
[tex]Vx = 1350cos(25\°) + 19.21(10.20)[/tex]
[tex]Vx = 1419.46\ m/s[/tex]
Final position in y:
[tex]23,500sin(55\°)=V_0y(t) + 0.5a_yt^2[/tex]
Where [tex]V_{0y}[/tex] is the initial velocity in y
So:
[tex]23,500sin(55\°)=1350sin(25\°)(10.20) + 0.5a_y(10.20)^2[/tex]
We solve for [tex]a_y[/tex]
[tex]23,500sin(55\°)-1350sin(25\°)(10.20) = 0.5a_y(10.20)^2[/tex]
[tex]a_y=\frac{23,500sin(55\°)-1350sin(25\°)(10.20)}{0.5(10.20^2)}[/tex]
[tex]a_y=258.18\ m/s^2[/tex]
Y component of velocity
[tex]V_y = 1350sin(25\°) +a_y(t)[/tex]
Where [tex]a_y[/tex] is the acceleration in the direction y
And [tex]a_y=258.18\ m/s^2[/tex]
Therefore:
[tex]Vy = 1350sin(25\°) + 258.18(10.20)[/tex]
[tex]Vy = 3203.97\ m/s[/tex]
Note then that the final velocity V is a two component vector
[tex]V = Vx+Vy[/tex]
[tex]V = 1419.46\ \^x+3203.97\ \^y\ \ m/s[/tex]
The magnitude of the final speed is:
[tex]|V| =\sqrt{V_x^2+V_y^2}\\\\|V|=\sqrt{(1419.46)^2+(3203.97)^2}\\\\|V|=3504.32\ m/s[/tex]
And the angle that forms with the x axis is:
[tex]\alpha=cos^{-1}(\frac{V_x}{|V|})\\\\\\\alpha =cos^{-1}(\frac{1419.46}{3504.32})\\\\\alpha=66\°[/tex]
