If after 2 seconds with v0y = 0 and s_y0 = 0, what is s_y given g = 9.8 m/s^2?

If after 2 seconds with v0y = 0 and s_y0 = 0, what is s_y given g = 9.8 m/s^2?

• A. -19.6 m
• B. -39.2 m
• C. 0
• D. 19.6 m

Answer: (A)

Vertical motion under constant downward acceleration follows s = s0 + v0 t - (1/2) g t^2 (taking upward as positive). With s0 = 0, v0 = 0, t = 2 s, and g = 9.8 m/s^2, the displacement is s = 0 + 0 - (1/2)(9.8)(2)^2 = -19.6 m. The negative sign shows the object has moved downward from the starting point, by 19.6 meters. If you drop the 1/2 factor, you’d get -39.2 m; if you flip the sign, you’d get +19.6 m; and if gravity didn’t affect it, you’d get 0. The result -19.6 m matches the expected downward displacement after 2 seconds.

Vertical motion under constant downward acceleration follows s = s0 + v0 t - (1/2) g t^2 (taking upward as positive). With s0 = 0, v0 = 0, t = 2 s, and g = 9.8 m/s^2, the displacement is s = 0 + 0 - (1/2)(9.8)(2)^2 = -19.6 m. The negative sign shows the object has moved downward from the starting point, by 19.6 meters. If you drop the 1/2 factor, you’d get -39.2 m; if you flip the sign, you’d get +19.6 m; and if gravity didn’t affect it, you’d get 0. The result -19.6 m matches the expected downward displacement after 2 seconds.