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
Sep 29, 2015

Answer:

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#

enter image source here

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#