Problem
based on Gravitation
1. A stone is thrown from the roof
of building in upward direction with velocity 9.8 m/sec.
a). Find maximum height.
b). Calculate total time taken to
reach the earth.
Sol. Given u = 9.8
m/sec v = 0 g = -9.8 m/sec^2
a). v^2 = u^2 + 2 g h
0 = (9.8)^2 - 2 * 9.8 * h
2 * 9..8 h = 9.8 * 9.8
h = 4.9 m
v = u + g t
0 = 9.8 - 9.8 t
9.8 t = 9.8
t = 1 sec
Total time is taken = 1 + 1 = 2 sec
2. A stone is thrown from the roof
of building of height 9.8m in downward direction.
a). Find final velocity.
b). Calculate total time taken to
reach the earth.
Sol. Given u = 0 m/sec g
= 9.8 m/sec^2 h = 9.8 m
a). v^2 = u^2 + 2 g h
v^2 = 0 + 2 * 9.8 * 9.8
v^2 = 9.8^2 * 2
v = 9.8 * √2
v = 9.8 * 1.414
v = 13.86 m/sec^2
v = u + g t
t = ( v- u ) / g
t = ( 13.86 – 0 ) / 9.8
t = 1.4 sec
 |
| science-ncert-solution |
3. A stone is thrown from the roof
of building in upward direction with velocity 4.9 m/sec.
a). Find maximum height.
b). Calculate total time taken to
reach the earth.
Sol. Given u = 4.9
m/sec v = 0 g = -9.8 m/sec^2
a). v^2 = u^2 + 2 g h
0 = (4.9)^2 - 2 * 9.8 * h
2 * 9..8 h = 4.9 * 4.9
h = 4.9/4 m
h = 1.225 m
v = u + g t
0 = 4.9 - 9.8 t
9.8 t = 4.9
t = 1/2 sec
Total time is taken = 1/2 + 1/2 = 1
sec
4. A stone is thrown from the roof
of building of height 4.9 m in downward direction.
a). Find final velocity.
b). Calculate total time taken to
reach the earth.
Sol. Given u = 0 m/sec g
= 9.8 m/sec^2 h = 4.9 m
a). v^2 = u^2 + 2 g h
v^2 = 0 + 2 * 9.8 * 4.9
v^2 = 9.8 * 9.8
v = 9.8 m/sec^2
b). v = u + g t
t = ( v- u ) / g
t = ( 9.8 – 0 ) / 9.8
t = 1 sec
5. Find the gravitational force
between earth and moon?
Mass of earth = 6 * 10^24 Kg
Mass of moon = 7.35 * 10^22 Kg
Distance between moon and earth =
3.84 * 10^8 m
Sol. Given Mass of earth = 6 * 10^24 Kg
Mass of
moon = 7.35 * 10^22 Kg
Distance
between moon and earth = 3.84 * 10^8 m
G = 6.7 *
10^11 N m^2 Kg^2
We know
that
F = (G m1
m2)/ R^2
Where m1 =
mass of earth
m2 = mass
of moon
F = (6.7 *
10^11) * (6 * 10^24) * (7.35 * 10^22) / (3.84 * 10^8)
F = (295.47
/ 3.84) * 10^41
F = 76.95
* 10^41
 |
| science-ncert-solution |
6. A ball is thrown from a height of
98 m downward and another ball is thrown upward at same time with velocity 39.2
/sec. Find the meeting point of both the balls.
Sol.
Given u = 0 g = 9.8
We know that
a). S1 = u t + ½ g t^2
S1 = 0 + ½ * 9.8 * t^2
S1 = 4.9 t^2 ….(1)
b). S2 = u t + ½ * g * t^2
S2 = 39.2t - ½ * 9.8 t^2
S2 = 39..2t - 4.9t^2 ….(2)
Given S1 + S2 = 98 ….(3)
By equation 1,2 and 3
We can write
4.9t^2 + 39.2t - 4.9t^2 = 98
39.2t = 98
t = 2.5 sec
By equation (1)
S1 = 4.9 * 2.5^2
S1 = 4.9 * 6.25
S1 = 30.625 m
S2 = 98 – S1
S2 = 98 – 30.625
S2 = 67.375 m
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