Question

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.

Answer 1

I found 6s and 46m

Explanation:

The ball starts with initial velocity $v_i=30m/s$ and it reaches maximum height where the velocity will be zero, $v_f=0$ During the upwards bit the acceleration of gravity $g=9.8m/s^2$ 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:

$v_f=v_i+at$

in our case:

0=30−9.8t 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:
$t=\frac{30}{9.8}=3s$
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:
$t_{tot}=2t=2.3=6$
The maximum height can be found considering:
$y_f-y_i=v_{i}t+\frac{1}{2}a{t}^2$
with our data:
$y_f-0=30.3+\frac{1}{2}9.8\times 3^2$
again the acceleration of gravity is directed downwards and we use the time to go up.
So:
$y_f=46cm$