If we ignore air resistance (as we normally do) and launch a projectile at an angle between zero and 90o, how does the magnitude (size) of the vertical component of the velocity change as time elapses? Group of answer choices

First, remember that:

Acceleration = dV/dt

where V is velocity and t is time.

When you launch a projectile (θ between 0° and 90° means that you are not launching it into the ground) you are applying an initial velocity v0, that can be separated in components, where the vertical component will be:

v0*sin(v)

That initial vertical velocity is a constant, so it does not depend on time.

Then when we differentiate with respect to the time, that constant will vanish, and we will get:

rate of change of the vertical velocity = vertical acceleration.

And as you know, when we ignore the air resistance the only force acting on the vertical axis is the gravitational force, such that the vertical acceleration will be:

a(t) = -9.8m/s^2

Where the negative sign is because this force points down.

Then the rate of change of the vertical velocity will be exactly that, -9.8m/s^2.

And the vertical velocity equation, to see exactly how the vertical velocity depends on time, will be:

V(t) = (-9.8m/s^2)*t + v0*sin(θ)

the correct choice is a .

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