A coin is dropped from a height of 750 feet. The height, s (ft), at time t (sec), is given by #s=-16t^2+750#. How long does it take for the coin to hit the ground?

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& what is the velocity when the coin hits the ground?

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Answer 1 · 2016-11-01T23:41:53

#6.85#sec

At ground level, #s=0#

# s=0 => -16t^2+750 = 0 #
# :. 16t^2 = 750 #
# :. t^2 = 750/16 #
# :. t^2 = 375/8 #
# :. t = +-sqrt(375/8) #
# :. t = +-6.85 # (2dp)

We need # t>0 #, and so #t=6.85#s (2dp)