First slide

Instantaneous acceleration

Question

A body starts from rest and falls vertically from a height of 19.6 metre. If g = 9.8 ms^-2, then the distance travelled by the body in the last 0.1 seconds of its motion is

Easy

A

1.9 metre
B

0.049 metre
C

17.7 metre
D

19. 6 metre

Solution

Time of free fall $= \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 19 .6}{9 .8}} = 2 \sec$
distance travelled in
$(2 - 0 .1) \sec = 1 .9 \sec is \frac{1}{2} s (1 .9)^{2}$
distance travelled in last 0.1 sec.
$\frac{1}{2} g (2)^{2} - \frac{1}{2} g (1 .9)^{2} = 1 .95 m$

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