# 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?

Jan 14, 2016

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 $23 m {s}^{-} 1$ and the acceleration due to gravity of $9.8 m {s}^{-} 2$ are in the same direction. We can use the following formula, but it will yield a quadratic equation we need to solve:

$d = u t + \frac{1}{2} a {t}^{2}$

Substituting in what we know, we get:

$9 = 23 t + \frac{1}{2} 9.8 {t}^{2}$

Rearrange to standard form for a quadratic equation:

$\frac{1}{2} 9.8 {t}^{2} + 23 t - 9 = 0$

I won't do it here, but I'd solve this with the quadratic formula, though there are other approaches. The roots are $t = 0.36 s \mathmr{and} - 5.06 s$. 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 ($23 m {s}^{-} 1 =$ about $83 k m {h}^{-} h$) 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!)