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?

Question

When will the ball hit the ground?

13482 views around the world

You can reuse this answer
Creative Commons License

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.