Motion in a Straight Line - Result Question 48

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50. A body dropped from top of a tower fall through $40 m$ during the last two seconds of its fall. The height of tower is $(g=10 m / s^{2})$

======= ####50. A body dropped from top of a tower fall through $40 m$ during the last two seconds of its fall. The height of tower is $(g=10 m / s^{2})$

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/motion-in-a-straight-line/motion-in-a-straight-line—result-question-48.md (a) $60 m$

(b) $45 m$

(c) $80 m$

(d) $50 m$

[1991]

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Answer:

Correct Answer: 50. (b)

Solution:

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

$D_n=u+\frac{a}{2}(2 n-1)$ we have, total distance travelled in last 2 seconds of fall is

$D=D_t+D _{(t-1)}$

$=[0+\frac{g}{2}(2 n-1)]+[0+\frac{g}{2}{2(n-1)-1}]$

$=\frac{g}{2}(2 n-1)+\frac{g}{2}(2 n-3)=\frac{g}{2}(4 n-4)$

$=\frac{10}{2} \times 4(n-1)$

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

Distance travelled in $t$ seconds is

where, $t=3 sec$

$s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 10 \times 3^{2}=45 m$