An object is thrown vertically from a height of 2 m at 18 m/s. How long will it take for the object to hit the ground?

Mar 19, 2018

The time taken for the object to hit the ground is $= 3.78 s$

Explanation:

Apply the equation of motion

$s = u t + \frac{1}{2} a {t}^{2}$

The initial velocity is $u = 18 m {s}^{-} 1$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Resolving in the vertical direction ${\uparrow}^{+}$

The distance travelled is $s = - 2 m$

Therefore,

$- 2 = 18 t - \frac{1}{2} \cdot 9.8 \cdot {t}^{2}$

Solving this quadratic equation for $t$

$- 2 = 18 t - 4.9 \cdot {t}^{2}$

$4.9 {t}^{2} - 18 t - 2 = 0$

$a = 4.9$

$b = - 18$

$c = - 2$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(- 18\right)}^{2} - 4 \left(4.9\right) \left(- 2\right) = 363.2$

Therefore,

$t = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{18 \pm \sqrt{363.2}}{2 \cdot 4.9}$

${t}_{1} = \frac{18 + 19.06}{9.8} = 3.78 s$

${t}_{2} = \frac{18 - 19.06}{9.8} = - 0.11$

Therefore,

The time taken for the object to hit the ground is $= 3.78 s$