Question

An object is thrown vertically from a height of `15 m` at `6 m/s`. How long will it take for the object to hit the ground?

Answer 1

2.4 seconds

Explanation:

If `t_1` is the time taken by object to reach same height then

`t_1 = (2u)/g = (2 × 6 ms^-1) / (9.8 ms^-2) = 1.2 s`

Velocity of object when it hits the ground is

`v = sqrt(u^2 + 2aS)`

`v = sqrt((6 ms^-1)^2 + (2 × 9.8 ms^-2 × 15m))`

`v = 18.2 ms^-1`

If `t_2` is further time taken by object to reach the ground then

`t_2 = (v - u)/g = (18.2 ms^-1 - 6 ms^-1)/(9.8ms^-2) = 1.2 s`

Total time taken `= t_1 + t_2 = 1.2s + 1.2s = 2.4s`

Answer 2

The time is about 2.5 s.

Explanation:

Compared to your first answer, this is an alternate method of solving.

I will choose to have up be the positive direction. Use the kinematic (suvat) formula
`s = u*t + 1/2 * a*t^2`
where

• `s = -15 m` because it will hit the ground - below the point it was thrown from
• `u = 6 m/s` positive because it was thrown up
• `a = -g = -9.8 m/s^2`.

Now, plugging in the data and solving for t
`-15 m = 6 m/s*t + 1/2*(-9.8 m/s^2)*t^2`

Adjust so that all 3 are on one side and zero is on the other (preparing for use of the quadratic equation)

`1/2*9.8 m/s^2*t^2 - 6 m/s*t -15 m = 0`

`4.9 m/s^2*t^2 - 6 m/s*t -15 m = 0`

`t = (6 +-sqrt(6^2 - 4 * 4.9 * (-15)))/(2*4.9)`

`t = (6 +-sqrt(36 + 294))/(9.8)`

`t = (6 +-sqrt(330))/(9.8) = (6 +-18.17)/(9.8)`

The - part of the `+-` will yield a negative time, so that solution is invalid. I will continue using the + part of the `+-`.

`t = (24.17)/(9.8) = 2.47 s`

I hope this helps,
Steve