Ncert Exercises & Solutions

Q.20 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 at 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/ $s^{2}$ )

Hint: Time taken in horizontal and vertical motion is the same.

Step 1: Find the time taken in the vertical motion.

Given,
The height difference between the two buildings = 9m
and g = 10m/ $s^{2}$
Let the man jumps from point A and lands on the roof of the next building at point B.
Taking motion in a vertical direction,

$\begin{array}{l} y = ut + \frac{1}{2} {at}^{2} \\ 9 = 0 \times t + \frac{1}{2} \times 10 \times t^{2} \\ 9 = 5 t^{2} \\ or t = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} \end{array}$

Step 2: Find the horizontal distance covered in the above time.

the horizontal speed of the man ( $u_{x}$ ) = 9m/s

$∴$ Horizontal distance travelled $= u_{x} \times t = 9 \times \frac{3}{\sqrt{5}} = \frac{27}{\sqrt{5}} m$ = 12 m

Step 3: Find whether will be able to land on the next building or not

Given, the horizontal distance between the two buildings = 10m

Now as the Horizontal distance traveled by the man is greater than 10 m, therefore, he will land on the next building.

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