motion under gravity

Motion Under Gravity When an object falls to the ground under the action of gravity, experiment shows that the object has a constant or uniform acceleration of about 980 cms" 2 , while it is falling (see p. 49). In SI units this is 9-8 ms-2 or 10 ms-2 approximately. The numerical value of this acceleration is usually denoted by the symbol g. Suppose that an object is dropped from a height of 20 m above the ground. Then the initial velocity u = 0, and the acceleration a = g = 10 m s -2 (approx). Substituting ins = ut+jat2 , the distance fallen s in metres is calculated from S = \gt2 = 5t2 . When the object reaches the ground, s = 20 m. .'. 20 = 5t2 , or t = 2 s Thus the object takes 2 seconds to reach the ground. If a cricket-ball is thrown vertically upwards, it slows down owing to the attraction of the earth. The ball is thus retarded. The magnitude of the retardation is 9-8 m s -2 , or g. Mathematically, a retardation can be regarded as a negative acceleration in the direction along which the object is moving; and hence a = — 9-8 m s~ 2 in this case. Suppose the ball was thrown straight up with an initial velocity, u, of 30 ms-1 . The time taken to reach the top of its motion can be obtained from the equation v = u + at. The velocity, v, at the top is zero; and since u = 30 m and a = —9-8 or 10 ms"2 (approx), we have = 30-10r. . 30 - •< = Io = 3s�The highest distance reached is thus given by s = ut+^at2 = 30x3-5x32 = 45 m.