Motion in a Straight Line - Result Question 48

50. A body dropped from top of a tower fall through 40m during the last two seconds of its fall. The height of tower is (g=10m/s2)

(a) 60m

(b) 45m

(c) 80m

(d) 50m

[1991]

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Answer:

Correct Answer: 50. (b)

Solution:

  1. (b) Let the body fall through the height of tower in nth seconds. From,

Dn=u+a2(2n1) we have, total distance travelled in last 2 seconds of fall is

D=Dt+D(t1)

=[0+g2(2n1)]+[0+g22(n1)1]

=g2(2n1)+g2(2n3)=g2(4n4)

=102×4(n1)

or, 40=20(n1) or n=2+1=3s

Distance travelled in t seconds is

where, t=3sec

s=ut+12at2=0+12×10×32=45m



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