Question

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?

Answer 1

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

Explanation:

Apply the equation of motion

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

The initial velocity is `u=18ms^-1`

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

Resolving in the vertical direction `uarr^+`

The distance travelled is `s=-2m`

Therefore,

`-2=18t-1/2*9.8*t^2`

Solving this quadratic equation for `t`

`-2=18t-4.9*t^2`

`4.9t^2-18t-2=0`

`a=4.9`

`b=-18`

`c=-2`

The discriminant is

`Delta=b^2-4ac=(-18)^2-4(4.9)(-2)=363.2`

Therefore,

`t=(-b+-sqrtDelta)/(2a)=(18+-sqrt(363.2))/(2*4.9)`

`t_1=(18+19.06)/9.8=3.78s`

`t_2=(18-19.06)/9.8=-0.11`

Therefore,

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