MCQ
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]
- ✓$4.9$
- B$9.8$
- C$12.4$
- D$19.6$
✓
Answer
Correct option: A.
$4.9$
a
(a)
(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.
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