# Can someone please explain how to do this problem?

## Cadets graduating from military school usually toss their hats high into the air at the end of the ceremony. One cadet threw his hat so that its distance d( t )= $- 16 {t}^{2}$ + 30t + 6. a. Find the distance above the ground of the hat 1 second after it was thrown. b. Find the time it took that hat to hit the ground. Give the exact time and a one-decimal-place approximation.

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#### Explanation

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#### Explanation:

I want someone to double check my answer

2
Nov 19, 2017

a. $20$ $m$.
b. $2.06$ $s$.

#### Explanation:

To find the height you only have to substitute $t$ in the equation for the number of seconds you what to know its possition,
d(t)=-16·1^2+30·1+6=20 $m$

and to find the time it took the hat to hit the ground you need to think a bit. If the hat is in the ground it means that de height is $0$, so you can equal the equation to $0$ and find the value of $t$.
$- 16 {t}^{2} + 30 t + 6 = 0$

we can use the formula for quadratics equations.
t=(-b+-sqrt(b^2-4ac))/(2a)=(-30+-sqrt(30^2-4·(-16)·6))/(2·(-16))=
$= \frac{- 30 \pm \sqrt{900 + 384}}{- 32} = \frac{- 30 \pm \sqrt{1284}}{- 32}$

$\cancel{{t}_{1} = \frac{- 30 + \sqrt{1284}}{- 32} = - 0.18 s}$
This solution is not possible because time can't be negative.

${t}_{2} = \frac{- 30 - \sqrt{1284}}{- 32} = 2.06$ $s$

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