9th std sci-gravitation_ppt.17-18

[13] A ball is thrown vertically upwards with a velocity of 49 m/s calculate
[1] maximum height to which it rises.
According to the equation of motion under gravity v2 – u2 = 2as
Where u = initial velocity of the ball
v= final velocity of the ball
s = Height achieved by the ball =
g = acceleration due to gravity = 9.8ms – 2 .
0 – 492 = 2 x 9.8 x h and h =
[2 ]the total time it takes to return to the surface of the earth Answer :
at maximum height, final velocity of the ball zero, i.e., v = 0 , U = 49 m/s
g = during upward motion g = − 9.8ms – 2 .
v= 19.6 Let ‘h’ be the maximum height attained by the ball , Hence
Let ‘t’ be the time taken by the ball to reach the height 122.5m,
then according to the equation of motion
Let v = u + gt we get 0 = 49 + t x (− 9.8)
− 9.8t = 49
t = 49/ 9.8 = 5sBut time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10s

[14] A stone is released from the top of a tower of height 19.6m Calculate its final velocity just before
Touching the ground.
Where u = initial velocity of the stone
v= final velocity of the stone
s = Height of the stone = 19.6m
g = acceleration due to gravity = 9.8ms – 2 .
There fore v2 – u2 = 2as
There fore v2 – 02 = 2 x 9.8 x 19.6
v2 = 2 x 9.8 x 19.6 = (19.6)2
v= 19.6 ms – 1

[15]A stone is thrown vertically upwards with an initial velocity of 40m/s . Taking g = 10m/s2, find the
Maximum height reached by the stone. What is the net displacement and the total distance covered by
the stone.
Where u = initial velocity of the stone = 40m/s
v = final velocity of the stone = 0
s = Height of the stone = 0
g = acceleration due to gravity = – 10ms – 2 .
There fore v2 – u2 = 2as
There fore 02 – 402 = 2 x h x (– 10 )
h = 40 x 40/20 = 80 m

16. Calculate the force of gravitation between the earth and the sun , given that the mass of the earth
= 6 x 1024 kg. And of the sun 2 x 10 30 kg the average distance between the two is 1.5 x 10 11m.
Answer: According to question, MSun = mass of the sun = = 6 x 1030 kg.
MEarth = mass of the Earth = = 6 x 1024 kg.
R =Average distance between the Earth and
Sun = 1.5 x 1011m.
R =From universal Law of gravitation,
Therefore, putting all the values
given in question we get
F = 6.67 x 10 – 11 x
F = 3.56 x 10 22 N
= 6.67 x 10 – 11 x
= 6.67 x 10 – 11 x
working
= 6.67 x 10 – 11 x
= 6.67 x 10- 11 x 5.33 x 10 54 – 22.
= 6.67 x 10 - 11 x 5.33 x 10 32.
= 6.67 x 5.33 x 10 32+(-11) .
= 35.5511 x 10 21.
= 3.56 x 10 22.
Force of
gravitation,
Earth
Sun

[12] Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth.
What is the weight in Newton's of 10kg object on the moon and on the earth?
Answer Weight of an object on the moon = 1/6 x weight of an object on the earth , also
Weight = Mass x Acceleration
Acceleration due to gravity g = 9 .8m/s2
Therefore weight of a 10kg object on the earth = 10 x 9.8 = 98N
And, weight of the same object on the moon = 1.6 x 9.8 = 16.3 N
[11] Why will a sheet of paper fall slower than one that is crumpled into a ball ?
Answer
When a sheet of paper is crumpled into a ball then its density increases. Hence resistance to
its motion through the air decreases and it falls faster than the sheet of paper.

GRAVITATION 2017 – 18
[17]A stone is allowed to fall from the top of a tower 100m high and at the same another stone is projected
vertically upwards from the ground with a velocity of 25m/s . Calculate when and where the two stones will
meet.
Let ‘t’ be the point at which two stones meet and let ‘h’ be their
height from the ground. It is given in the question that height of
the tower is H = 100m Now first consider the stone which falls
from the top of the tower. So distance covered by this stone at
time ‘t’ can be calculated use equation of motion.
x – x0 = u0t + ½ gt2.
Since initial velocity u = o so we get
100 – x = ½ gt2 .----------[1]
the distance covered by the same stone that is thrown
The upward direction from ground is
X = u 0t − ½ gt2
In this case initial velocity is 25m/s . So,
x = 25t − ½ gt2 . --------[2]
Stone falling
Stone vertically up
wards
x = 21.6m

Adding equation [1] and [2] we get,
100 = 25t
or t = 4s
Putting value in equation [2]
In x = 25t − ½ gt2 .
x = 25 x 4 − ½ x 9.8 x (4)2 .
x = 100 − ½ x 9.8 x 16.
x = 100 − 9.8 x 8.
x = 100 − 78.4
x = 21.6
Stone falling
Stone vertically up
wards

[18] A ball thrown up vertically returns to the thrower after 6s. Find
Answer
[a] The velocity with which it was thrown up
[b] The maximum height it reaches, and
[c] Its position after 4s.
[a] Time of ascent is equal to the time of descent. The ball takes a total of 6s for its upward and
Downward journey , Hence it has taken 3s to attain the maximum height. Final velocity of the at maximum
V = u + gt will give
at the maximum height v = 0 . Acceleration due to gravity g = − 9.8ms-2 . Equation of motion
0 = u +(− 9.8 x 3 )
u = 9.8 x 3 = 29.4 ms −1
Hence , the ball was thrown upwards with a velocity of 29.4 ms −1
[b] let the maximum height attained by the ball be ‘h’ initial velocity during the upward journey
u =29.4 ms −1 . Final velocity , v = 0, Acceleration due to gravity g = − 9.8ms-2 From the equation of motion
S = ut + ½ at2.
h = 29.4 x 3 + ½ x 9.8 x (3)2 = 44.1 m

[c] Ball attains the maximum height after 3s after attaining this height , it will start falling downwards.
In this case, Initial velocity u = 0, position of the ball after 4 s of the throw is given by the distance travelled
by it during its downward journey in 4 s – 3s = 1 s
Equation of motion
S = ut + ½ gt2. will give
S = 0 x t + ½ x 9.8 x 12.
S = 4.9 x 1.
S = 4.9
Total height = 44.1m
This means that the ball is 39.2m ( 44.1m – 4 .9m)
Above the ground after 4 s

9th std sci-gravitation_ppt.17-18

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9th std sci-gravitation_ppt.17-18