Motion in a Straight Line - Result Question 41
43. If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
(a) $(u+g t) t$
(b) $u t$
(c) $\frac{1}{2} g t^{2}$
(d) $u t-\frac{1}{2} g t^{2}$
[2003]
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Answer:
Correct Answer: 43. (c)
Solution:
- (c) Let bodytakes $T$ sec to reach maximum height.
Then $v=u-g T$
$v=0$, at highest point.
So, $T=\frac{u}{g}$
Velocity attained by body
in $(T-t) \sec v=u-g(T-t)$
$v=u-g T+g t=u-g \frac{u}{g}+g t \quad(\because T=u / g)$
or $v=g t$
$\therefore$ Distance travelled in last $t sec$ of its ascent
$s=v t-\frac{1}{2} g t^{2}$
$s=(g t) t-\frac{1}{2} g t^{2}=\frac{1}{2} g t^{2}$
If a body is projected vertically upward, then, final velocity, $v=0$; intial speed, $u=g t$ and height attained, $h=\frac{u^{2}}{2 g}=\frac{g^{2} t^{2}}{2 g}=\frac{1}{2} g t^{2}$
