Motion in a Straight Line - Result Question 48
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 $(g=10 m / s^{2})$
(a) $60 m$
(b) $45 m$
(c) $80 m$
(d) $50 m$
[1991]
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Answer:
Correct Answer: 50. (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, $t=3 sec$
$s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 10 \times 3^{2}=45 m$
