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A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. - Physics

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प्रश्न

A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building? (take g = 10 m/s2)

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उत्तर

Horizontal Projectile: When a body is projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers an equal distance in the horizontal direction in equal intervals of time.

Time of flight: If a body is projected horizontally from a height h with velocity u and the time taken by the body to reach the ground is T, then

`h = 0 + 1/2 gT^2`  ......(For vertical motion)

`T = sqrt((2h)/g)`

Horizontal range: Let R be the horizontal distance travelled by the body

`R = uT + 1/2 0T^2`  .......(For horizontal motion a = 0)

`R = u sqrt((2h)/g)`

We will apply kinematic one by one downward and horizontal. We first consider motion horizontal and there is no horizontal force which can affect horizontal motion. The horizontal velocity therefore remains constant and so the object covers an equal distance in the horizontal direction in equal intervals of time.

According to the problem, the horizontal speed of the man (ux) = 9 m/s Horizontal distance between the two buildings = 10 m

Height difference between the two buildings = 9 m and g = 10 m/s2


Let the man jumps from point A and land on the roof of the next building at point B.

Taking motion in the vertical direction,

`y = ut + 1/2 at^2`

9 = `0 xx t + 1/2 xx 10 xx t^2`

9 = 5t2

or t = `sqrt(9/5) = 3/sqrt(5)`

∴ Horizontal distance travelled = `u_x xx t = 9 xx 3/sqrt(5) = 27/sqrt(5)` m ≈ 12 m

The horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.

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अध्याय 3: Motion In a Straight Line - Exercises [पृष्ठ १७]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 3 Motion In a Straight Line
Exercises | Q 3.20 | पृष्ठ १७

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