A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air resistance.]

H = 40 m, the height from which the ball is dropped.

m = 1 kg, the mass of the ball

Assume g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity is zero.

If t = the time of flight, then

40 m = (1/2) * g * (t s) ² = 0.5*9.8*t²

t² = 40/4.9 = 8.1633

t = 2.857 s

Answer: 2.9 s (nearest tenth)

2.856s Distance traveled under constant acceleration as a function of time is given by x = (gt^2) / 2 where g is the acceleration and t is time. In this case acceleration is due to gravity and is 9.81m/s^2. The distance of interest is x=40m. Substituting and solving 40 = (9.81*t^2) / 2 80 = 9.81*t^2 8.1549 = t^2 2.856s = t