9th Class (CBSE) Science: Gravitation

Question: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Answer:

According to the equation of motion under gravity:
v² − u² = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s^-2
∴ v² − 0² = 2 × 9.8 × 19.6
v² = 2 × 9.8 × 19.6 = (19.6)²
v = 19.6 m s^-1
Hence, the velocity of the stone just before touching the ground is 19.6 ms^-1_.

Question: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer:

According to the equation of motion under gravity:
v² − u² = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = −10 m s⁻²
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)2 = 2×h×(-10)
h = (40×40)/20 = 80 m

Therefore, total distance covered by the stone during its upward and downward
journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (−80) = 0

Question: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Answer:

According to question,

M_S = Mass of the Sun = 2×10³⁰ kg
M_E = Mass of the Earth = 6×10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5×10¹¹ m
From Universal law of gravitation,

Question: A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Answer:

Let t be the point at which two stones meet and let h be their height from the ground.

Height of the tower is H = 100 m (Given)

It is clear from the question that we need to calculate time when the two stones met. After calculating time, we will also be able to calculate the distance.
Now, first consider the stone which falls from the top of the tower.

Initial velocity (u) = 0

So, distance covered by this stone at time t can be calculated using equation of motion