#### Answer

a). For free fall, $v=u+at=0-gt=-9.8t\, m/s$ , we consider downwards to be negative. Graph is as shown.
b). For free fall, $S=ut+\frac{1}{2}at^{2}=0-0.5\times 9.8t^{2}=-4.9t^{2}$
So, $y=-4.9t^{2}$. Graph is as shown.

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Published by
Pearson

ISBN 10:
0-32160-183-1

ISBN 13:
978-0-32160-183-4

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