Force and Acceleration
F = mg
The gravitational force on an object equals its mass times the acceleration due to gravity.
Value of g
g = GM/R² = 9.8 m/s²
Where G is the universal gravitational constant, M is Earth's mass, and R is Earth's radius.
Equations of Motion
v = u + gt
s = ut + ½gt²
v² = u² + 2gs
These equations describe motion under constant acceleration due to gravity.
Example 9.2 (from NCERT): A car falls off a ledge and drops to the ground in 0.5 s.
Let g = 10 m/s² (for simplifying calculations)
Find: (i) Speed on striking the ground
(ii) Average speed during the 0.5 s
(iii) Height of the ledge from the ground
Given Values:
Time, t = 0.5 s
Initial velocity, u = 0 m/s (dropped, not thrown)
Acceleration due to gravity, g = 10 m/s²
Acceleration of the car, a = +10 m/s² (downward)
(i) Speed on striking ground:
v = u + at
v = 0 + 10 × 0.5
v = 5 m/s
(ii) Average speed:
Average speed = (u + v)/2
= (0 + 5)/2
= 2.5 m/s
(iii) Height of ledge:
s = ut + ½at²
s = 0 × 0.5 + ½ × 10 × (0.5)²
s = ½ × 10 × 0.25
s = 1.25 m
(i) Speed on striking ground = 5 m/s
(ii) Average speed = 2.5 m/s
(iii) Height of ledge = 1.25 m
Example 9.3 (from NCERT): An object is thrown vertically upwards and rises to a height of 10 m.
Calculate: (i) The velocity with which the object was thrown upwards
(ii) The time taken by the object to reach the highest point
Given Values:
Distance travelled, s = 10 m
Final velocity, v = 0 m/s (at highest point)
Acceleration due to gravity, g = 9.8 m/s²
Acceleration, a = -9.8 m/s² (upward motion, opposing gravity)
(i) Initial velocity:
Using v² = u² + 2as
0 = u² + 2 × (-9.8) × 10
-u² = -2 × 9.8 × 10
u² = 196
u = 14 m/s
(ii) Time taken:
Using v = u + at
0 = 14 + (-9.8) × t
9.8t = 14
t = 14/9.8 = 1.43 s
(i) Initial velocity, u = 14 m/s
(ii) Time taken, t = 1.43 s