# An object is thrown vertically from a height of 12 m at  1 m/s. How long will it take for the object to hit the ground?

time $t = 1.67028567 \text{ }$seconds

#### Explanation:

$y = {v}_{y} t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$- 12 = 1 \cdot t + \frac{1}{2} \cdot - 9.8 \cdot {t}^{2}$

$- 24 = 2 t - 9.8 {t}^{2}$

$9.8 {t}^{2} - 2 t - 24 = 0$

$4.9 {t}^{2} - t - 12 = 0$

$t = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(4.9\right) \left(- 12\right)}}{2 \cdot \left(4.9\right)}$

$t = \frac{+ 1 \pm \sqrt{1 - 4 \left(4.9\right) \left(- 12\right)}}{2 \cdot \left(4.9\right)}$

$t = 1.67028567 \text{ }$seconds

God bless...I hope the explanation is useful.

Apr 30, 2016

$2$ seconds

#### Explanation:

Assuming that the object is thrown upwards and then falls down, we can use one of the kinematic equations to solve for the time it takes for the object to hit the ground.

For the given problem, we will let positive values represent a downward direction.

$\Delta d = 12 m$

${v}_{i} = - 1 \frac{m}{s}$

$a = 9.81 \frac{m}{s} ^ 2$

Deltat=?

Use the kinematic equation, $\Delta d = {v}_{i} \Delta t + \frac{1}{2} a \Delta {t}^{2}$, to solve for $\Delta t$.

$\Delta d = {v}_{i} \Delta t + \frac{1}{2} a \Delta {t}^{2}$

$0 = \frac{1}{2} a \Delta {t}^{2} + {v}_{i} \Delta t - \Delta d$

Plug in the values.

$0 = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\frac{1}{2} \left(9.81 \frac{m}{s} ^ 2\right)} \Delta {t}^{2} + \left(\textcolor{t e a l}{- 1 \frac{m}{s}}\right) \Delta t$ $\textcolor{v i o \le t}{- 12 m}$

Use the quadratic formula to solve for $\Delta t$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{t e a l}{- 1 \frac{m}{s}}\right) \pm \sqrt{{\left(\textcolor{t e a l}{- 1 \frac{m}{s}}\right)}^{2} - 4 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\frac{1}{2} \left(9.81 \frac{m}{s} ^ 2\right)} \left(\textcolor{v i o \le t}{- 12 m}\right)\right)}}{2 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\frac{1}{2} \left(9.81 \frac{m}{s} ^ 2\right)}\right)}$

$x = 1.67 s \textcolor{w h i t e}{i} , \textcolor{w h i t e}{i} \textcolor{red}{\cancel{\textcolor{b l a c k}{- 1.47 s}}}$

Since we are looking for the time it takes for the object to hit the ground after it is thrown, we take the positive value, $1.67 s$, ignore $- 1.47 s$.

By rounding $1.67 s$ to $1$ significant digit, it becomes $2 s$, which is the time it takes for the object to hit the ground.