# An object is thrown vertically from a height of 6 m at  2 m/s. How long will it take for the object to hit the ground?

Feb 7, 2017

The object reaches the ground after 1.33 s.

#### Explanation:

The convenient thing about using the equations of motion for a problem like this, is that you do not have to split the motion into an "upward" part and a "downward" part. Just as long as the acceleration is constant (it is) and you keep track of the signs on your numbers, you can do it is a single calculation.

Here it is. Use this equation: $\Delta y = {v}_{o} \Delta t + \frac{1}{2} a \Delta {t}^{2}$

because you do not know the final velocity of the object.

In this problem, $\Delta y = - 6 m$ as the landing point is below the starting location. Also, the acceleration is $- 9.8 \frac{m}{s} ^ 2$, but ${v}_{o} = + 2 \frac{m}{s}$, as the object begins its motion in an upward direction. So,

$- 6 = 2 \Delta t - 4.9 \Delta {t}^{2}$

This is a quadratic equation! Write it in standard form $a {x}^{2} + b x + c = 0$ and use the quadratic formula.

$4.9 \Delta {t}^{2} - 2 \Delta t - 6 = 0$

$\Delta t = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(4.9\right) \left(- 6\right)}}{2 \left(4.9\right)}$

$\Delta t = \frac{2 \pm \sqrt{121.6}}{9.8}$

$\Delta t = \frac{2 \pm 11}{9.8}$

The two answers are $\Delta t = \frac{13}{9.8} = 1.33 s \mathmr{and}$Deltat=-9/9.8 = -0.92s

The first answer is the one we want. The second represents the time the object would have been on the ground if, instead of being thrown from 6 m, it had been thrown 0.92 s earlier, from the ground.