A ball is dropped from the top of a building 25m high. How much time will the ball need to reach the ground?

1 Answer
Steve M
Feb 1, 2018

# 2.2488 \ s \ (4 \ dp) #

Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #

If we consider the downward direction as positive, and zero relative to the initial position the ball will start with an initial velocity of #0 \ ms^-1#, and it will have a displacement of #25 \ m # and will fall under the effect of gravity alone.

# { (s=,25,m), (u=,0,ms^-1), (v=,"Not Required",ms^-1), (a=,g=9.8,ms^-2), (t=,T,s) :} #

So applying:

# s =ut+1/2at^2 #

We find that:

# 25 =0+1/2g t^2 #
# :. T^2 = 50/g#

Taking #9=9.8 \ ms^-2# this gives us:

# T^2 = 50/9.8 = 5.1020 \ (4 \ dp)#
# T = +- 2.2488 \ (4 \ dp) #

and we discard the negative solution

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