Question

An object is thrown vertically from a height of `15 m` at `19 m/s`. How long will it take for the object to hit the ground?

Answer 1

`4.55 s`

Explanation:

The vertical distance the object travels is given by the equation
`y = v_(y0)t - 1/2g*t^2` where `v_(y0)` is the vertical component of the initial velocity, `g` is the acceleration due to gravity and `t` is the time taken to reach the highest point (where the velocity is zero).
The time can be found from the equation `v_(yf) = v_(yo) - g*t` by using the fact that `v_(yf) = 0`

Thus `0=19-9.8*t`
`t = 19/9.8 = 1.94 s`

The height the object reaches is therefore `19t - 9.8*t^2/2`
`=18.4 m`

We now have to work out how long the object will take to fall `18.4+15 = 33.4 m`

`33.4 =0*t + 1/2g*t^2`
`t = sqrt(33.4*2/9.8) =2.61 s`

The total time is `1.94 + 2.61 = 4.55 s`