An object is thrown vertically from a height of 2 m at 3 m/s. How long will it take for the object to hit the ground?

Apr 25, 2016

I found $1.01 s$

Explanation:

I would try using the general kinematic relationship:
${y}_{f} - {y}_{i} = {v}_{i} t + \frac{1}{2} a {t}^{2}$
where:
${y}_{i}$ is the initial position;
${y}_{f}$ is the final position (the ground);
${v}_{i}$ is the initial velocity;
$a$ is the acceleration of gravity (directed downwards):
so you should get:
$0 - 2 = 3 t - \frac{1}{2} \cdot 9.8 {t}^{2}$
$4.9 {t}^{2} - 3 t - 2 = 0$
solving the Quadratic Equation gives us:
${t}_{1} , 2 = \frac{3 \pm \sqrt{9 + 39.2}}{9.8} =$
two solutions:
${t}_{1} = 1.01 s$
${t}_{2} = - 0.4 s$