An object is thrown vertically from a height of 9 m at  6 m/s. How long will it take for the object to hit the ground?

Mar 29, 2018

The time to hit the ground is $= 2.099 s$

Explanation:

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

The initial velocity is $u = 6 m {s}^{-} 1$

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

The distance is $s = - 9 m$

Applying the equation of motion

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

$- 9 = 6 t - \frac{9.8}{2} {t}^{2}$

$4.9 {t}^{2} - 6 t - 9 = 0$

Solving this quadratic equation in $t$

$t = \frac{\left(6\right) \pm \sqrt{36 + 4 \cdot 9 \cdot 4.9}}{2 \cdot 4.9}$

$= \frac{6 \pm \sqrt{212.4}}{9.8}$

${t}_{1} = \frac{6 + 14.57}{9.8} = 2.099 s$

${t}_{2} = \frac{6 - 14.57}{9.8} = - 0.87 s$

The time to hit the ground is $= 2.099 s$

graph{4.9x^2-6x-9 [-9.99, 10.02, -5, 4.995]}

Mar 29, 2018

First, calculate the time the object takes to reach the maximum height.

To calculate the time taken to reach the maximum height, you have the following info:

color(teal)(u=6m/s, v=0,a=-9.8m/s^2 ($a = - g$, as the object here move upwards.)

According to the first equation of motion,

color(teal)(v=u+at.

Plug in the values:
color(teal)(0=6-9.8t
or, color(teal)(9.8t=6
or, color(teal)(t=6/9.8sapprox0.61s

To reach the same height from where it was thrown (i. e., $9 m$ above the ground), the object will take the same time.

Thus, time taken to go up and come down to the $9 m$ level, total time taken
color(teal)(=0.61s+0.61s=1.22s

Now, all you need is the time taken by the object to go down the last color(blue)(9m

For that,
color(blue)(S=9m,u=6m/s,a=+9.8m/s^2

Use the third equation to get:
color(blue)(v^2=u^2+2aS

Plug in the values; color(blue)(vapprox14.57

Now, find the time using the first equation of motion:
color(blue)(14.57=6+9.8t
color(blue)(tapprox0.87

Thus, total time,
color(green)(T=(0.87+1.22)s=2.09s