A ball is thrown straight up from the top of a 80 foot tall building with an initial speed of 64 feet per second. The height of the ball as a function of time can be modeled by the function h(t)=-16t^2+64t+80. When will the ball be 144 feet above ground?

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When will the ball hit the ground?

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Answer 1 · 2017-10-26T14:41:16

after #2# seconds

#h(t) = -16t^2+64t+80#

factorise:

#h(t) = 16(-t^2+4t+5)#

in this example, #16(-t^2+4t+5) = 144#

#-t^2+4t+5 = 144/16 = 9#

#-t^2+4t+5 = 9#

#-t^2+4t-4 = 0#

#t = (-b +- sqrt(b^2-4ac))/(2a)#

#= (-4 +- sqrt(16-16))/(-2)#

#=(-4 +-0)/-2#

#=-4/-2# (repeated root)

#=2#

#t = 2#

assuming the time is in seconds, it will take #2# seconds for the ball to reach #144# feet above ground.

desmos.com/calculator

red: #y=-x^2+4x+5#
blue: #y=9#

to find where #-t^2+4t+5=9#, the point of intersection of both graphs is used - #(2,9)#

[#x# in the function is #t# time]

this means that after #2# seconds, the ball will have reached #9*16# feet (see factorisation), or #144# feet.