In this question we have given
velocity of missile=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 the final velocity of missile we will first find the acceleration required
Let x be the horizontal component of distance
x - vertical 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
horizontally: 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
Therefore
V = √(V'x² + V'y²) = 3504 m/s
therefore magnitude of final velocity of missile=3504m/s
