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

final velocity = 3504 m/s  

Explanation:

Given data:

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

To find:

Final velocity: ?

Formula:

x = Vx*t + ½*ax*t²  

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

Solution:

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

similarly vertically:

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  

V'y = Vy + ay*t

     = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

V = √(V'x² + V'y²)

   = 3504 m/s  

the rearrangement of particles in a physical change. hope this