Which equation describes vertical velocity as a function of time in projectile motion near Earth's surface, taking upward as positive?

Which equation describes vertical velocity as a function of time in projectile motion near Earth's surface, taking upward as positive?

• A. V_y = V0y + g t
• B. V_y = V0y - g t
• C. V_y = -V0y + g t
• D. V_y = V0y - (1/2) g t^2

Answer: (B)

When gravity is constant and upward is taken as positive, the vertical acceleration is downward and equal to -g. With constant acceleration, velocity changes linearly in time: v_y(t) = v_y0 + a_y t. Substituting a_y = -g gives v_y(t) = V0y - g t. This form matches the initial condition that at time zero the vertical velocity is V0y, and it shows that the upward velocity decreases by g each second due to gravity, eventually reaching zero at the top of the trajectory and becoming negative as the object falls. The other expressions don’t fit the situation: one would imply gravity is acting upward, which is incorrect; another would flip the initial sign of the velocity; and the last uses t^2, which is appropriate for displacement, not velocity (the displacement is y = y0 + V0y t - (1/2) g t^2).

When gravity is constant and upward is taken as positive, the vertical acceleration is downward and equal to -g. With constant acceleration, velocity changes linearly in time: v_y(t) = v_y0 + a_y t. Substituting a_y = -g gives v_y(t) = V0y - g t.

This form matches the initial condition that at time zero the vertical velocity is V0y, and it shows that the upward velocity decreases by g each second due to gravity, eventually reaching zero at the top of the trajectory and becoming negative as the object falls.

The other expressions don’t fit the situation: one would imply gravity is acting upward, which is incorrect; another would flip the initial sign of the velocity; and the last uses t^2, which is appropriate for displacement, not velocity (the displacement is y = y0 + V0y t - (1/2) g t^2).