Question

A healthy young man 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 at (1.8 m)?Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

solution:
For Kid:
Initial velocity=u=0 m/s
acceleration=a=g=9.8m/s2
Distance= s=11.8 -1.8=10m
Now s=ut +1/2at²
10=0+1/2x9.8 xt²
t²=10/4.9=2.04 sec
t=1.42 sec

In this time the man has to reach at the bottom of the building 
Velocity=v=s/t=7/1.42
=4.9m/s
∴The man should run at a speed of 4.2 m/s to catch the child.

Answer 2

Answer:

Over 1.8 m from ground the kid should be catched.

For kid,

initial velocity u = 0

Acceleration = 9.8 m/s2

Distance S = 11.8 – 1.8 = 10 m

S = ut + ½ at2

⇒ 10 = 0 + ½ (9.8) t2

⇒ t2 = 2.04 ⇒ t = 1.42.

In this time the man has to reach at the bottom of the building.

Velocity s/t = 7/1.42 = 4.9 m/s