a-stone-is-launched-vertically-upward-from-the-top-of-a-building-with-initial-ve

Question

A stone is launched vertically upward from the top of a building with initial velocity v0=15 $$v_0 = 15  m/s$$	ext{m/s} and initial position s0=45 $$s_0 = 45  m$$	ext{m}. Find the position and velocity functions for all times when the stone is above ground level (s=0 s = 0 ). Use g=9.8 m/s2g=9.8	ext{ m/s}^2.

Accepted Answer

Position: s(t)=45+15t−4.9t2s(t)=$$45+15t-4.9t^2$$Velocity: v(t)=15−9.8tv(t)=15-9.8t

Answer

Position: s(t)=45−15t+4.9t2s(t)=$$45-15t+4.9t^2$$Velocity: v(t)=15−9.8tv(t)=15-9.8t

Answer

Position: s(t)=45+15t−4.9t2s(t)=$$45+15t-4.9t^2$$Velocity: v(t)=15−4.9tv(t)=15-4.9t

Answer

Position: s(t)=45+15t−9.8t2s(t)=$$45+15t-9.8t^2$$Velocity: v(t)=15+9.8tv(t)=15+9.8t