If a stone is tossed from the top of a 270 meter building, the height of the stone as a function of time is given by h(t) = -9.8t2 – 10t + 270, where t is in seconds, and height is in meters. After how many seconds will the stone hit the ground?

1 Answer
George C.
Jun 28, 2015

Solve #h(t) = 0# using the quadratic formula to get:

#t = (10+-sqrt(10^2-(4xx-9.8xx270)))/(2xx-9.8)#

#t~=4.76# or #t~=-5.78#

Discard the negative solution to get #t ~= 4.76# seconds.

Explanation:

#h(t)# is of the form #at^2+bt+c#, with #a=-9.8#, #b=10# and #c=270#

The roots of #h(t) = 0# are given by the formula:

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

#= (10+-sqrt(10^2-(4xx-9.8xx270)))/(2xx-9.8)#

#=-(10+-sqrt(10684))/19.6#

#~=-(10+-103.36)/19.6#

#t~=4.76# or #t~=-5.78#

Discard the negative solution to get #t ~= 4.76# seconds.

The negative solution relates to a prequel to the story in which the stone is thrown up from the ground #5.78# seconds before it passes the top of the building on the way back down.

Related questions
See all questions in Linear, Exponential, and Quadratic Models
Impact of this question
1443 views around the world
You can reuse this answer
Creative Commons License