Answer to Question #291783 in Classical Mechanics for Tai

Question

A bullet leaves the barrel of a gun at a speed or 320 m/s at an angle of 45 degrees above the horizontal. At the moment the bullet leaves the barrel, the acceleration of the bullet is:

A. 0

B. 320m/s to the right

C. 9.8m/s upward

D. 9.8m/s downward

Answer 1

$a=\frac{\Delta v}{\Delta t}=\frac{dv}{dt}=0,$

answer A.