A 3.00-kilogram mass is thrown vertically upward

with an initial speed of 9.80 meters per second.

What is the maximum height this object will

reach? [Neglect friction.]

(1) 1.00 m (3) 9.80 m

(2) 4.90 m (4) 19.6 m

2) 4.90 m. Using Newton's equation of motion, we have, vf^2 = vi^2 + 2gs where a is the acceleration due to gravity, vf the final velocity (0 when it reaches maximum height), vi the initial velocity (9.80) and s the distance travelled. Substituting the values, we have 0 = 9.80^2 + 2 (-9.81) s. Rearranging, s = (9.80^2) / 2 (9.81) = 4.90 m