Question

a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

Answer 1

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