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.

1 Answer
Gió
Sep 29, 2015

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=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=6s

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The maximum height can be found considering:
y_f-y_i=v_it+1/2at^2
with our data:
y_f-0=30*3-1/2(9.8)(3)^2
again the acceleration of gravity is directed downwards and we use the time to go up.
So:
y_f=46m

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