# Motion of an Object Under Gravity

The g is constant near the surface of earth. Therefore, all the equations for uniformly accelerated motion of bodies become valid when acceleration a is replaced by g.

Modified equations of motion are:

v = u + gt

s = ut + ½gt2

v2 = u2 + 2gs

where u and v are the initial and final velocities and s is the distance covered in time t.

Example 1: Take the mass of the earth to be 6 ×1024 kg and its radius as 6.4 × 106 m. Calculate the value of g. (G = 6.7 × 10–11 N m2 kg–2).

g = GM/R2

= 9.8 ms–2

Example 2: The mass of the earth is 6 ×1024 kg and that of the moon is 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84 × 108 m, calculate the force exerted by the earth on the moon. G = 6.7 × 10–11 N m2 kg–2.

The mass of the earth, m1 = 6 × 1024 kg

The mass of the moon, m2 = 7.4 ×1022 kg

The distance between the earth and the moon, r = 3.84 × 108 m

G = 6.7 × 10–11 N m2 kg–2

The force exerted by the earth on the moon is

F = Gm1m2/r2

F = 2.01 × 1020 N

Example 3: A ball is thrown vertically upwards and rises to a height of 122.5 m. Calculate

1. the velocity with which the ball was thrown upwards
2. the time taken by the ball to reach the highest point

Take g = 9.8 ms–2

Distance travelled, s = 122.5 m

Final velocity, v = 0 ms–1

Acceleration due to gravity, g = 9.8 ms–2

(i) From equation: v2 = u2 + 2gs

0 = u2 + 2(–9.8 ms–2) × 122.5 m

For upward motion g is taken as negative.

–u2 = –2 × 9.8 × 122.5 m2s–2

u2 = 2401 m2s–2

u = 49 ms–1

Thus, the velocity with which the ball was thrown upwards is 49 ms–1.

(ii) From equation: v = u + gt

0 = 49 ms–1 + (9.8 ms–2) × t

Therefore, t = 49/9.8 = 5s