Answer:
final velocity = 3504 m/s
Explanation:
<em>Given data:</em>
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
<em>To find:</em>
Final velocity: ?
<em>Formula:</em>
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
<em>Solution:</em>
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
<em>similarly vertically:</em>
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
