Question

A healthy youngman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height (1.8 m)?

Answer 1

At the speed of 4.92 m/s, he should run to catch the kid.

Explanation:

Step 1:

For Children ,

                    u is initial velocity  

                    a = g = is acceleration  

                    s is distance  

                    t is time  

                     v is velocity  

                     u = 0 m/s

                     a = g = 9.8 m/s²

                     s = 11.8 - 1.8 = 10m

Step 2:

Now,                $s=u t+\frac{1}{2} a t^{2}$

                      $10=0 \times t+\frac{1}{2} \times 9.8 \times t^{2}$

                      $10=0+\frac{1}{2} \times 9.8 t^{2}$

                      $10=\frac{1}{2} \times 9.8 t^{2}$

                      $20=9.8 t^{2}$

                       $t^{2}=\frac{20}{9.8}$

                       $t^{2}=2.04$

                       $t=\sqrt{2.04}$

                       $t=1.42 second$

Step 3:

The guy will enter in this time at the bottom of the building ,

             Velocity  $=\mathrm{v}=\frac{\text { distance }}{\text { time }}$

                               $v=\frac{7}{1.42}$  (distance = 7 m)

                                  = 4.92 m/s

Thus, to catch the child the man will run at a pace of 4.92 m / s.            

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

4.92 m/s of speed required