If a rose thrown from a platform has a height, h(t), given by the formulah(t)= -16t^2 + 32t + 20, how can you use the quadratic formula to determined how far it will travel before hitting ground?

Oct 26, 2015

$116 m$

Explanation:

The quadratic function $h \left(t\right)$ gives the height of the rose at any time t after being released.
Since $h \left(0\right) = 20$ we conclude that the original height of the rose when thrown at time $t = 0$ is $20 m$, ie the height of the platform is 20m above ground.

We now seek to determine at what time t will $h \left(t\right) = 0$ as this will be the time when the rose is at the ground.

This is equivalent to solving the quadratic equation
$- 16 {t}^{2} + 32 t + 20 = 0$

We may use the quadratic formula to solve this and obtain

$t = \frac{- 32 \pm \sqrt{{32}^{2} - 4 \times - 16 \times 20}}{2 \times - 16}$

$= \frac{- 32 \pm 48}{-} 32$

$= - 0 , 5 \mathmr{and} 2 , 5$

But since time is scalar it cannot be negative so we discard the -0,5 value and conclude that $t = 2 , 5 s$

This means that the ball will be at height zero, ie the ground, exactly 2,5 seconds after being thrown.

We now still need to find what the maximum height reached by the stone is and to do that we differentiate the height function and set it to zero.

$\frac{\mathrm{dh}}{\mathrm{dt}} = 0 \iff - 32 t + 32 = 0 \iff t = 1 s$

We can now evaluate the height function at the time when t = 1 second to find the distance above ground after the ascent to yield

$h \left(1\right) = - 16 {\left(1\right)}^{2} + 32 \left(1\right) + 20 = 68 m$

Thus, the total distance travelled by the object is $2 \times \left(68 - 20\right) + 20 = 116 m$.