Question

Francesca throws a ball. The height in metres, h, of the ball above the ground t seconds after she throws it is given by h = 1 + 14t - 5t². How many seconds after Francesca throws the ball does it hit the ground? Give your answer to 3 significant figures.

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Answer 1

Check the picture below, so Francesca's throw looks more or less like that, and it hits the ground when h(t) = 0

`\stackrel{h}{0}=1+14t-5t^2\implies 5t^2-14t-1=0 \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-14}t\stackrel{\stackrel{c}{\downarrow }}{-1}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`t= \cfrac{ - (-14) \pm \sqrt { (-14)^2 -4(5)(-1)}}{2(5)} \implies t = \cfrac{ 14 \pm \sqrt { 196 +20}}{ 10 } \\\\\\ t= \cfrac{ 14 \pm \sqrt { 216 }}{ 10 }\implies t=\cfrac{ 14 \pm 6\sqrt { 6 }}{ 10 }\implies t=\cfrac{ 7 \pm 3\sqrt { 6 }}{ 5 }\implies t\approx \begin{cases} ~~ 2.870~\checkmark\\ -0.070 \end{cases}`

so there are two valid values, however we can't use the negative one, since the phenomena is for seconds and we can't have negative seconds in this case.