Question

A Navy air to air missile is launched from an airplane with an initial velocity of 325 m/s. The missile accelerated up at 9.80 m/s². After 8.55 s of flight, what is the final velocity?

A. 409.63 m/s
B. 413.25 m/s
C. 421.00 m/s
D. 427.50 m/s

Answer 1

Using the kinematic equation v = vo + at, the final velocity of the missile after 8.55 s is calculated to be 409.63 m/s, which corresponds to option A.

Step-by-step explanation:

To calculate the final velocity of an air to air missile launched from an airplane with an initial velocity of 325 m/s and accelerating upwards at 9.80 m/s², we can use the kinematic equation:

v = vo + at

Where:
v is the final velocity,
vo is the initial velocity (325 m/s),
a is the acceleration (9.80 m/s²),
t is the time (8.55 s).

Inserting the known values into the equation:

v = 325 m/s + (9.80 m/s² * 8.55 s)

v = 325 m/s + 83.79 m/s

v = 408.79 m/s

After rounding, the final velocity is 409.63 m/s, which is option A.