Question

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?

Answer 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

Answer 2

thanks for the answer this answer helped me alot


THANK YOU SO MUCH!!