Question
A point object is thrown vertically upwards at such a speed that it returns to the thrower after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed time graph for the object and use it to find the distance travelled by it in the last second of its journey.

Answer

$\text{v}=\text{u}+\text{at}$ with $\text{v}=-\text{u},\text{a}=-9.8\text{m/s}^2$ $\text{t}=6\text{s}$ $\therefore \text{u}=29.4\text{m/s}$ Time of ascent + Time of descent = 6s $\text{t}_1+\text{t}_2=6\text{s}$ and $\text{t}_1=\text{t}_2=\text{t}$ $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$ $\text{t}=3\text{s}$ $\text{s}=(29.4\text{m/s})(3\text{s})+\frac{1}{2}(-9.8\text{m/s}^2)(3)^2$ $=88.2\text{m}-44.1\text{m}$ $=44.1\text{m}$ Distance travelled by it in last second = area under the curve = 24.5m.

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