Question

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?

Answer 1

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

Explanation:

Resolve in the vertical direction `uarr^+`

The initial velocity is `u=6ms^-1`

The acceleration due to gravity is `g=-9.8ms^-2`

The distance is `s=-9m`

Applying the equation of motion

`s=ut+1/2at^2`

`-9=6t-9.8/2t^2`

`4.9t^2-6t-9=0`

Solving this quadratic equation in `t`

`t=((6)+-sqrt(36+4*9*4.9))/(2*4.9)`

`=(6+-sqrt(212.4))/9.8`

`t_1=(6+14.57)/9.8=2.099s`

`t_2=(6-14.57)/9.8=-0.87s`

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

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

Answer 2

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., `9m` above the ground), the object will take the same time.

Thus, time taken to go up and come down to the `9m` 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`

hope that helped...Thanks for reading!!!