# Question #af942

Mar 13, 2017

$t = 3.6 s$, rounded to one decimal place.

#### Explanation:

Using FPS system of units.

The kinematic expression is
$h \left(t\right) = {h}_{0} + u t + \frac{1}{2} g {t}^{2}$
Inserting given values and noting that acceleration due to gravity$32 f t \cdot {s}^{-} 2$ is acting against the direction of motion we get

$h \left(t\right) = 7 + 54 t + \frac{1}{2} \left(- 32\right) {t}^{2}$
$h \left(t\right) = 7 + 54 t - 16 {t}^{2}$
Which is the given equation.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
We can start solution from here.

When the ball reaches ground $h = - 7 f t$
Inserting in the given height equation we get
$- 7 = 7 + 54 t - 16 {t}^{2}$
$\implies 16 {t}^{2} - 54 t - 14 = 0$
$\implies 8 {t}^{2} - 27 t - 7 = 0$

Solving the quadratic with the inbuilt graphic tool we get two roots as below. Ignoring the $- v e$ root as time can not be negative we get

$t = 3.6 s$, rounded to one decimal place.