Answer:
V=[tex]3204m/s[/tex]
Direction is normal to the incline
Explanation:
This question tests on projectile motion.
First, calculate the acceleration of the missile.
Incline=55º,Distance=23500m, t=10seconds.
Horizontal motion of projectile
[tex]x=V_xt+0.5a_xt\\23000Cos55\textdegree=1350Cos25\textdegree\times10.20+0.5a_x(10.20)^2\\a_x=19.2m/s^2[/tex]
Final velocity
[tex](V_x)\prime=V_x+a_xt=1350Cos25\textdegree+19.2\times10.20=1419m/s[/tex]
The vertical motion of the missile can be calculated as:
[tex]23500 Sin55.0\textdegree =1350Sin25\textdegree\times10.2+0.5a_y\times10.2^2\\a_y=258.2m/s^2[/tex]
Final Velocity is:
[tex](V_y)\prime=V_y+a_yt=1350Sin25\textdegree+258.2\times10.20=3204m/s[/tex]
Combining both we get
[tex]V=\sqrt(V\prime x^2+V\primey^2)=3504.2m/s[/tex]
*Misile's motion is normal to the 55º incline.
