Question #ccb40

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Answer 1 · 2017-03-04T14:07:54

Explanation below:
- 10m
- 2sec

Start by setting $t = 0 \Rightarrow h = 10$ $t = 0 implies h = 10$ , so the diver is 10m above the water line to start with.

When the diver hits the waterline, we can say that:

$h (t) = 0$ $h(t) = 0$

$\Rightarrow - 5 t^{2} + 5 t + 10 = 0$ $implies -5t^2 + 5t + 10 = 0$

$\Rightarrow t^{2} - t - 2 = 0$ $implies t^2 - t -2 = 0$

Which we can factor as:

$(t - 2) (t + 1) = 0 \Rightarrow t = - 1, 2$ $(t - 2)(t+1) = 0 implies t = -1, 2$

The negative solution is nonsense, so the diver hits the waterline at $t = 2$ $t = 2$ s.