Question

A healthy young man standing at a distance of 7m from a 10m high building see a kid slipping from the top floor. With speed (assumed uniform) should he run to catch the kid at the arms height (1.8m)? PLEASE GIVE EXACT ANSWER IN DECIMALS FOR SAME VALUES IF CORRECT I FOLLOW YOU I MARK AS BRAINLEST

Answer 1

Answer:

At point B (i.e. over 1.8 m from ground) the kid should be catched. For kid initial velocity u = 0 Acceleration = 9.8 m/s2 Distance S = 10– 1.8 = 8.2m S = ut + ½ at2 ⇒ 8.2= 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.

Please mark my answer as brainliest

Answer 2

For the child:-

•Initial velocity=0

•Acceleration due to gravity(g)=9.8m/s²

•Distance(s)=10-1.8=8.2m

Now,by using 2nd equation of motion,when u=0,we get----

=>s=1/2gt²

=>8.2=1/2(9.8)t²

=>8.2=4.9t²

=>t²=8.2/4.9

=>t²=1.67

=>t=1.29s

Now,in this time the man has to reach the bottom of the building.

=>Speed=Distance/time

=>7/1.29

=>5.42m/s

Thus,the man should run with a speed of 5.42m/s to catch the kid.