A bullet is fired with a muzzle velocity of 1446 ft/sec from a gun aimed at an angle of 5 degrees above the horizontal. Find the vertical component of the velocity.

126.03 ft/sec

Explanation:

From the question above,

V₁ = VsinФ Equation 1

Where V₁ = vertical component of the velocity, V = Velocity acting on the x-y plane, Ф = angle to the horizontal.

Given: V = 1446 ft/sec, Ф = 5°

Substitute into equation 1

V₁ = 1446sin(5)

V₁  = 1446(0.0872)

V₁  = 126.03 ft/sec.

Hence the vertical component of the velocity = 126.03 ft/sec

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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real, smaller and upside-down.   so the answer should be b.