Question

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 isof a lower height than the first. if his speed is $9m/s$, the (horizontal) distance between the two building is $10m$ and the height difference is$9m$, will he be able to land on the next building $(Takeg=10m/s^2)$

Answer 1

Step:1
Draw rough figure for the given problem.


Step:2 Find time taken by the man to fall vertically downwards.

Formula used: $h=ut+\frac{1}{2}g{t}^2$

Given,

horizontal speed of the man,$u_x=9m/s$

Horizontal distance between the two building , $x=10m$

Height difference between the two building $h=9m$

Let the man jump from first building to land on the next buliding at a point B.

Now , for motion in the vertical direction Vertical speeed of the man, $u=0$

Acceleration in vertical direction,

$a=g=10m/s^2$

$h=ut+\frac{1}{2}g{t}^2$

$9=0\times t+\frac{1}{2}\times 10\times t^2$

$9=5t^2$

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

Step3: Find horizontal distance travelled by the man.

Horizontal distance travelled by the man

$u_x\times t$

$9x\times \sqrt{\frac{9}{5}}$

$\approx 12m$

As the horizontal distance travelled by the man is greater than the distance between the two building i.e.$10m$. so, the man will land on the next building

Final answer: Yes