Question

A missile is moving 1350 m/s at a 25.0 degree angle. It needs to hit a target 23500 m away in a 55.0 degree direction in 10.20s. What is the direction of its final velocity?

Answer 1

V=
`3204m/s`

Direction is normal to the incline

Step-by-step explanation:

This question tests on projectile motion.

First, calculate the acceleration of the missile.

Incline=55º,Distance=23500m, t=10seconds.

Horizontal motion of projectile

`x=V_xt+0.5a_xt\\23000Cos55\textdegree=1350Cos25\textdegree*10.20+0.5a_x(10.20)^2\\a_x=19.2m/s^2`

Final velocity

`(V_x)\prime=V_x+a_xt=1350Cos25\textdegree+19.2*10.20=1419m/s`

The vertical motion of the missile can be calculated as:

`23500 Sin55.0\textdegree =1350Sin25\textdegree*10.2+0.5a_y*10.2^2\\a_y=258.2m/s^2`

Final Velocity is:

`(V_y)\prime=V_y+a_yt=1350Sin25\textdegree+258.2*10.20=3204m/s`

Combining both we get

`V=\sqrt(V\prime x^2+V\primey^2)=3504.2m/s`

*Misile's motion is normal to the 55º incline.