Question

A ball is dropped from rest from the top of a building, which is 106 m high. The magnitude of the gravitational acceleration g = 9.8 m/s2 Keep 2 decimal places in all answers. In this problem, the following setup is convenient: Take the initial location of ball (the top of the building) as origin x0 = 0 Take DOWNWARD as +x (a) How far (in meters) does the ball fall in the first 3 s ?

Answer 1

Answer:

44.1 m

Explanation:

initial velocity of ball, u = 0

height of building, H = 106 m

g = 9.8 m/s^2

t = 3 second

Let the ball travels a distance of h in first 3 seconds.

Use second equation of motion

$s=ut+\frac{1}{2}at^{2}$

h = 0 + 0.5 x 9.8 x 3 x 3

h = 44.1 m

Thus, the distance traveled by the ball in first 3 seconds is 44.1 m.