Question

An object is thrown vertically from a height of `9 m` at `23 ms^-1`. How long will it take for the object to hit the ground?

Answer 1

Assuming the initial velocity is downward, the object will hit the ground after `0.36 s`.

Explanation:

The question is a little ambiguous: is the object initially thrown upward, downward, horizontally or at an angle? I will assume it is thrown downward, but as a bonus will calculate the time taken if it is thrown upward.

Assuming it is thrown downward, the initial velocity of `23 ms^-1` and the acceleration due to gravity of `9.8 ms^-2` are in the same direction. We can use the following formula, but it will yield a quadratic equation we need to solve:

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

Substituting in what we know, we get:

`9 = 23t + 1/2 9.8t^2`

Rearrange to standard form for a quadratic equation:

`1/2 9.8t^2 + 23t - 9 = 0`

I won't do it here, but I'd solve this with the quadratic formula, though there are other approaches. The roots are `t = 0.36 s and -5.06 s`. We'll ignore the negative root and state that it takes `0.36 s`. for the stone to reach the ground.

(as a quick check for sensibleness you could calculate that a stone simply dropped from 9 m would take `1.35 s` to hit the ground: since in this case it was thrown downward quite fast (`23 ms^-1 =` about `83 kmh^-h`) the smaller answer makes sense.)

(If we assume instead that the stone was thrown upward, we simply give the initial velocity a negative sign, since it is in the opposite direction to the acceleration. Working it through gives `5.06 s`. Hmm... that number seems oddly familiar!)