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:

  1. (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, $h=\frac{u^{2}}{2 g}=\frac{g^{2} t^{2}}{2 g}=\frac{1}{2} g t^{2}$



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