# A ball is launched vertically upwards from ground level with an initial velocity of 30 m/s. About how much time elapses before it reaches the ground? About what maximum altitude does it reach above ground? Please explain I'm so confused.

Sep 29, 2015

I found $6 s$ and $46 m$

#### Explanation:

The ball starts with initial velocity ${v}_{i} = 30 \frac{m}{s}$ and it reaches maximum height where the velocity will be zero, ${v}_{f} = 0$. During the upwards bit the acceleration of gravity $g = 9.8 \frac{m}{s} ^ 2$ is slowing it down up to the maximum height where the ball finally stops;

You can say that the final velocity depends upon the initial velocity AND the contribution of the acceleration that operates for a certain time $t$, or:

${v}_{f} = {v}_{i} + a t$

in our case:
$0 = 30 - 9.8 t$ the acceleration is negative because it is directed downwards (see the diagram).
So you have that the time to reach the maximum height will then be:
$t = \frac{30}{9.8} = 3 s$.

This is the time to go up; you double it to consider also the second leg of the trip when the ball comes back to the ground: so:

${t}_{\text{tot}} = 2 t = 2 \cdot 3 = 6 s$

The maximum height can be found considering:
${y}_{f} - {y}_{i} = {v}_{i} t + \frac{1}{2} a {t}^{2}$
with our data:
${y}_{f} - 0 = 30 \cdot 3 - \frac{1}{2} \left(9.8\right) {\left(3\right)}^{2}$
again the acceleration of gravity is directed downwards and we use the time to go up.
So:
${y}_{f} = 46 m$