Amissile 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?

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

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.

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