Motion In A Straight Line Ques 45
45. 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 $(g=10 m / s^{2})$
[1991]
(a) $60 $ $m$
(b) $45 $ $m$
(c) $80 $ $m$
(d) $50 $ $m$
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Answer:
Correct Answer: 45.(b)
Solution:
- (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 $ $m$
BETA
