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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.

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 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?

Answers (1)

In vertical motion,

u_{y} = 0

a = 10 m/s^{2}

s = 9m & t=t

now, s = u_{y}t+\frac{1}{2}at^{2}

thus, 9=0(t)+\frac{1}{2}(10)(t^{2})

Therefore,

t=\sqrt{\frac{9}{5}}

  =\frac{3}{\sqrt{5}}sec

Now, the horizontal distance covered by the person will be,

u_{x}\times\; t=9\left ( \frac{3}{\sqrt{5}} \right )

            =\frac{27}{\sqrt{5}}=\frac{27}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}

            =27\times\frac{2.236}{5}

            =12.07\; m

Therefore, the person will reach the building which s next farther the first edge by 12.7 - 10 = 2.07\; m.

 

Posted by

infoexpert22

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