Page 115 - Understanding NCERT Science 09

We know that an object experiences u +v
acceleration during free fall. From Eq. (9.9), (ii) average speed = 2
this acceleration experienced by an object is = (0 m s + 5 m s )/2
–1
–1
independent of its mass. This means that all = 2.5 m s –1
objects hollow or solid, big or small, should (iii) distance travelled, s = ½ a t 2
fall at the same rate. According to a story, = ½ × 10 m s × (0.5 s) 2
–2
Galileo dropped different objects from the top = ½ × 10 m s × 0.25 s 2
–2
of the Leaning Tower of Pisa in Italy to prove = 1.25 m
the same. Thus,
As g is constant near the earth, all the (i) its speed on striking the ground
equations for the uniformly accelerated = 5 m s –1
motion of objects become valid with (ii) its average speed during the 0.5 s
acceleration a replaced by g. = 2.5 m s –1
The equations are: (iii) height of the ledge from the ground
v = u + at (9.10)
= 1.25 m.
1
s = ut + at 2 (9.11)
2
2
2
v = u + 2as (9.12) Example 9.3 An object is thrown vertically
upwards and rises to a height of 10 m.
where u and v are the initial and final velocities Calculate (i) the velocity with which the
and s is the distance covered in time, t. object was thrown upwards and (ii) the
In applying these equations, we will take time taken by the object to reach the
acceleration, a to be positive when it is in the highest point.
direction of the velocity, that is, in the
direction of motion. The acceleration, a will Solution:
be taken as negative when it opposes the
motion. Distance travelled, s = 10 m
Final velocity, v = 0 m s –1
Acceleration due to gravity, g = 9.8 m s –2
Example 9.2 A car falls off a ledge and Acceleration of the object, a = –9.8 m s –2
drops to the ground in 0.5 s. Let (upward motion)
g = 10 m s –2 (for simplifying the 2 2
(i) v = u + 2a s
calculations). 0 = u + 2 × (–9.8 m s ) × 10 m
–2
2
(i) What is its speed on striking the –u = –2 × 9.8 × 10 m s
2
2 –2
ground? -1
(ii) What is its average speed during the u = 196 m s
0.5 s? u = 14 m s -1
(ii) v = u + a t
(iii) How high is the ledge from the
0 = 14 m s – 9.8 m s × t
–1
–2
ground?
t = 1.43 s.
Thus,
Solution:
–1
(i) Initial velocity, u = 14 m s , and
Time, t = ½ second (ii) Time taken, t = 1.43 s.
Initial velocity, u = 0 m s –1
Acceleration due to gravity, g = 10 m s –2
Acceleration of the car, a = + 10 m s –2 uestions
(downward) 1. What do you mean by free fall?
(i) speed v = a t Q 2. What do you mean by acceleration
–2
v = 10 m s × 0.5 s due to gravity?
= 5 m s –1

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