1.3 Solutions
17
As the time interval between the first and second drop is equal to that of the
second and the third drop (drops dripping at regular intervals), time taken by
the second drop is t 2 =
1
2
√
2
s; therefore, distance travelled by the second
drop is
S =
1
2
gt
2
2 =
1
2
× 9.8 ×
1
2
√
2
2
= 0.6125 m
1.11 Height h = area under the υ − t graph. Area above the t-axis is taken positive
and below the t-axis is taken negative. h = area of bigger triangle minus area
of smaller triangle.
Now the area of a triangle = base × altitude
h =
1
2
× 3 × 30 −
1
2
× 1 × 10 = 40 m
1.12 (a) Time for the ball to reach water t 1 =
2 h
g
=
2 × 4.9
9.8
= 1.0 s
Velocity of the ball acquired at that instant v = gt 1 = 9.8 × 1.0 =
9.8 m/s.
Time taken to reach the bottom of the lake from the water surface
t 2 = 5.0 − 1.0 = 4.0 s.
As the velocity of the ball in water is constant, depth of the lake,
d = vt 2 = 9.8 × 4 = 39.2 m.
(b) < v >=
total displacement
total time
=
4.9 + 39.2
5.0
= 8.82 m/s
1.13 For the first stone time t 1 =
2 h
g
=
2 × 44.1
9.8
= 3.0 s.
Second stone takes t 2 = 3.0 − 1.0 = 2.0 s to strike the water
h = ut 2 +
1
2
gt
2
2
Using h = 44.1 m, t 2 = 2.0 s and g = 9.8 m/s 2 , we find u = 12.25 m/s
1.14 Transit time for the single journey = 0.5 s.
When the ball moves up, let υ 0 be its velocity at the bottom of the window, v 1
at the top of the window and v 2 = 0 at height h above the top of the window
(Fig. 1.15)
17
As the time interval between the first and second drop is equal to that of the
second and the third drop (drops dripping at regular intervals), time taken by
the second drop is t 2 =
1
2
√
2
s; therefore, distance travelled by the second
drop is
S =
1
2
gt
2
2 =
1
2
× 9.8 ×
1
2
√
2
2
= 0.6125 m
1.11 Height h = area under the υ − t graph. Area above the t-axis is taken positive
and below the t-axis is taken negative. h = area of bigger triangle minus area
of smaller triangle.
Now the area of a triangle = base × altitude
h =
1
2
× 3 × 30 −
1
2
× 1 × 10 = 40 m
1.12 (a) Time for the ball to reach water t 1 =
2 h
g
=
2 × 4.9
9.8
= 1.0 s
Velocity of the ball acquired at that instant v = gt 1 = 9.8 × 1.0 =
9.8 m/s.
Time taken to reach the bottom of the lake from the water surface
t 2 = 5.0 − 1.0 = 4.0 s.
As the velocity of the ball in water is constant, depth of the lake,
d = vt 2 = 9.8 × 4 = 39.2 m.
(b) < v >=
total displacement
total time
=
4.9 + 39.2
5.0
= 8.82 m/s
1.13 For the first stone time t 1 =
2 h
g
=
2 × 44.1
9.8
= 3.0 s.
Second stone takes t 2 = 3.0 − 1.0 = 2.0 s to strike the water
h = ut 2 +
1
2
gt
2
2
Using h = 44.1 m, t 2 = 2.0 s and g = 9.8 m/s 2 , we find u = 12.25 m/s
1.14 Transit time for the single journey = 0.5 s.
When the ball moves up, let υ 0 be its velocity at the bottom of the window, v 1
at the top of the window and v 2 = 0 at height h above the top of the window
(Fig. 1.15)