# An object is thrown vertically from a height of 3 m at 18 m/s. How long will it take for the object to hit the ground?

Dec 30, 2015

0.159 seconds, when the object is thrown vertically downwards.
3.833 seconds, when the object is thrown vertically upwards.

#### Explanation:

I'll consider the upward direction to be positive and the downward direction to be negative.

The acceleration due to earth's gravity, $g = - 9.8 \frac{m}{s} ^ 2$

When the object is thrown downwards,

Its initial velocity, ${v}_{0} = - 18 \frac{m}{s}$

From Newton's Equations of Motion, we have:

$s = {v}_{0} t + \frac{1}{2} a {t}^{2}$

where $s$ is the displacement of the object, equal to $- 3 m$

${v}_{0}$ is the displacement of the object, equal to $- 18 \frac{m}{s}$

$a$ is the acceleration of the object, equal to $g = - 9.8 \frac{m}{s} ^ 2$

and $t$ is the time taken by the object to complete this motion.

Substituting the values in the above equation, we get:

$- 3 = - 18 t + \frac{1}{2} \cdot \left(- 9.8\right) \cdot {t}^{2}$

$\frac{1}{2} \cdot \left(9.8\right) \cdot {t}^{2} + 18 t - 3 = 0$

Solving the quadratic equation, you will get two values for $t$

$t = 0.159 s \mathmr{and} t = - 3.833 s$

Clearly, time cannot have a negative value. Therefore, the object will take 0.159 seconds to reach the ground.

Now, what if it was thrown vertically upwards?

It's simple! You just need to change the sign of ${v}_{0}$ to account for the change in direction!

This time around, the quadratic equation you will get will be:

$\frac{1}{2} \cdot \left(9.8\right) \cdot {t}^{2} - 18 t - 3 = 0$

Solving the quadratic equation, you will get two values for $t$

$t = - 0.159 s \mathmr{and} t = 3.833 s$

It's interesting to note that you will receive the exact same values as the last time, but with a reversal of signs.

Clearly, time cannot have a negative value. Therefore, the object will take 3.833 seconds to reach the ground.