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Derivation of Position-Velocity Relation by Graphical Method

A ball is thr...

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A ball is thrown vertically upward from the top of a building with velocity of 10 m/s. If it takes 4 seconds for the ball to reach ground level, what is the height of the building?

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Q. A ball is thrown vertically upwards from the top of a building of height

H .

It goes to a height of

20 m

and then comes back to the ground. Time taken by the ball to reach the maximum height from the top of the building is half of the time taken to reach the ground from the maximum height. What is the height of the building? (Take

g = 10 m s^{- 2}

)

Q. A person throws ball with a velocity

‘ v^{'}

from top of a building in vertically upward direction. The ball reaches the ground with a speed of

^{'} 3 v^{'}

. The height of the building is:

Q. A ball is thrown with a velocity of

20 m / s e c

from top of a multistory building the height of point from were the ball is thrown in

25 m

from the ground. (i) How height is the ball rise (ii) How long will it take before hitting the ground? Take

g = 10 m / s

Q. A ball is thrown vertically upwards with a velocity of 20 m

s^{- 1}

from the top of a multistorey building of 25 m high. How high will the ball rise from top of the building? (Take g = 10 m

s^{- 2}

A stone is dropped from a building top takes 4 seconds to reach the ground. What is the height of the building

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Derivation of Position-Velocity Relation by Graphical Method

Standard IX Physics

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