Question

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18. A ball thrown up vertically returns to the thrower after 6 s.
Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.​

Answer 1

Given:-

• Final velocity ,v = 0m/s

• Total time taken ,t = 6s

• Acceleration due to gravity, g = 10m/s²

To Find:-

(a) The velocity with which it was thrown up,

(b) The maximum height it reaches, and

(c) Its position after 4 s.

Solution:-

It is given that the total time taken by the ball is 6s

Its mean that the ball take 3 second to reach maximum height and 3 second to comes it initial position.

[a]

By using 1st equation of motion

• v = u+at

Substitute the value we get

Or , 0 = u + 10×3

Or, u = 30m/s.

Therefore, the initial velocity of the ball is 30m/s.

[b]

By using 2nd equation of motion

• h = ut + 1/2at²

Substitute the value we get

Or, h = 30×3 + 1/2×(-10) ×3²

Or, h = 90 - 5×9

Or, h = 90 -45

Or, h = 45m

Therefore ,the maximum height attained by the ball is 45m.

[c]

Firstly we need to calculate the distance travelled by ball in 1s . By using 2nd equation of motion

• s = ut +1/2at²

Substitute the value we get

s = 0×1 +1/2×(-10) ×1²

s = 0 + (-5)×1

s = -5 m

Therefore, the position of the ball after 4s is 45 -5 = 40m.

Answer 2

Answer:

Importance of The Universal Law of Gravitation

It binds us to the earth. It is responsible for the motion of the moon around the earth. It is responsible for the motion of planets around the Sun. Gravitational force of moon causes tides in seas on earth.