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
a=ΔvΔt=dvdt=0,a=\frac{\Delta v}{\Delta t}=\frac{dv}{dt}=0,a=ΔtΔv=dtdv=0,
answer A.
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