KVPY SA Exam 2016 Question Paper

Question : 27 of 80

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A stone thrown down with a seed u takes a time

t_1

to reach the ground , while another stone, thrown upwards from the same point with the same speed, takes time

t_2

. The maximum height the second stone reaches from the ground is

Solution:

-h=-Ut_1+{-1}/2gt_1^2

h= Ut_1+{ 1}/2gt_1^2.....(1)

-h=Ut_2-1/2gt_2^2...(1)

Ut_1+Ur_2=1/2gt_2^2-1/2gt_1^2

(from 1 & 2)

U(t_1+t_2)=g/2(t_2-t_1)×(t_1+t_2)

U=g/2(t_2-t_1)

∴ Max height

= h+U^2/{2g}

=Ut_1+1/2gt_1^2+U^2/{2g}

=g/2t_1(t_2-t_1)+1/2gt_1^2+1/{2g}g^2/4(t_2-t_1)^2

=(g/2)[ (t_2t_1-t_1^2)+t_1^2+(t_2-t_1)^2/4]

=(g/2)[ {4t_2t_1-4t_1^2+4t_1^2+t_2^2+t_1^2-2t_1t_2}/4]

=g/8(t_1+t_2)^2

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