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  4. A body is projected vertically upward with speed 40 m / s. The distance travelled by body in the last second of upward joumey is [take g=9.8 m / s 2 and neglect effect of air resistance]

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Q. A body is projected vertically upward with speed $40\, m / s$. The distance travelled by body in the last second of upward joumey is [take $g=9.8\, m / s ^{2}$ and neglect effect of air resistance]

Motion in a Straight Line

Solution:

As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second of descent.
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$t=1\, s$ ($1^{\text {st }}$ second)
$-x_{2}=u t-\frac{1}{2} g \times 1^{2}$
$x_{2}=\frac{1}{2} \times 9.8 \times 1^{2} (\because u=0)$
$\Rightarrow x_{2}=4.9\, m$
This distance is constant for every body thrown with any speed.