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Grade 12General Physics

A healthy youngman standing from a distance of 6m from a 11.5m high building sees a kid slipping from the top floor.With what uniform accleration in m/s2 (starting from rest) should he run to catch the kid in his arms height(1.5m)?Take g=10m/s2.

Profile image of Hansraj Gyanendra Singh Rajawat
10 Years agoGrade 12
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5 Answers

Profile image of Vikas TU
ApprovedApproved Tutor Answer10 Years ago
For perfect catching there should be a perfect timing ‘t’ .i.e. time taken by kid and the young man to catch should be same.
For kid,
11.5 – 1.5 = 0 + 0.5gt^2 
or
t^2 = 2 ….........................(1)
 
For young man,
6 = 0 + 0.5at^2
or
at^2 = 12 ….............................................(2)
put t^2 from eqn1. 1in eqn. 2
a*2 = 12
a = 6 m/s^2
Profile image of Hansraj Gyanendra Singh Rajawat
10 Years ago
 
11.5 – 1.5 = 0 + 0.5gt^2 
or 
and
6 = 0 + 0.5at^2 explain these steps??????????????
I didnot understand them.........................................
Profile image of Vikas TU
10 Years ago
Dear Hansraj,
eqn.
11.5 – 1.5 = 0 + 0.5gt^2  is described as:
distance covered by kid is 11.5 -1.5 = 10 because man would catch him from 1.5 above the ground.
thus the kid would not cover whole distance of 11.5 m distace.
that’s why we must subtract that 1.5 meter.
 
and initial velocity is zero for kid because he has fallen from rest.
g is gravity .i.e 10 m/s^2 and time taken is t to reach in man’s arms.
 
 
 
Profile image of Vikas TU
10 Years ago
Secondly,
in eqn.
6 = 0 + 0.5at^2
6 meter is the distance of young man to the building which he must cover in that same time ‘ t ‘ in order to catch him perfectly.
he runs from rest position therfore, u =0
and let a be the accln. we have to calculate and the time ‘ t ‘ is obvious.
Profile image of Hansraj Gyanendra Singh Rajawat
10 Years ago
ok thanks you cleared my all doubt that I am facing in this particular chapter so thank you again...................
 
 

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