# Algebra

Help Please, ty A ball is launched at 20 meters per second (m/s) from a 60 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 20t + 60, where s is in meters.  What is the height above the ground when the object is launched? How long before the object hits the ground after launch? What is the maximum height of the object?

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1. s = height above ground
s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)

at t = 0
s = 60 (clearly :)

now when does it hit the bleak earth?
That is when s = 0
4.9 t^2 - 20 t -60 = 0
solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)
t = 6.09 or - 2.01
use t = 6.09

now to do the last part there are two obvious ways to get t at the peak
1. look for vertex of parabola
2. look for halfway between t = -2.01 and t = 6.09
I will do it the hard (11) waay by completing the square
4.9 t^2 - 20 t = -(s-60)
t^2 - 4.08 t = -.204 s + 12.2
t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16

(t-2.04)^2 = -.204(s-80.2)
so
top at 80.2 meters at t = 2.04 s
===============
quick check on time
should be average of 6.09 and -2.01
=4.08 /2 = 2.04 check

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