A body is projected vertically upward with speed 40 ,m / s . distance travelled by body in last seco

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2.Motion in Straight Line

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

A

$4.9$

B

$9.8$

C

$12.4$

D

$19.6$

Solution

(a)

As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second of descent.

$t=1 s \left(1^{\text {st }} \text { second }\right)v=0$

$-x_2=u t-\frac{1}{2} g \times 1^2 x_1+x_2$

$x_2=\frac{1}{2} \times 9.8 \times 1^2(\because u=0)$

$\Rightarrow x_2=4.9 \,m$

This distance is constant for every body thrown with any speed.

Standard 11

Physics

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