An object is launched at 9.8 meters per second from a 73.5-meter tall platform. The object's height s (in meters) after t seconds is given by the equation s(t)= -4.9t(-4.9t) - 9.8t = 73.5. When does the object strike the ground?

Question

I can't figure out how to factor it fully. I got to -4.9(t^2 +2t +15) = 0

Answer 1

it was evidently launched straight up in which case:
h(t) = 73.5 + 9.8 t - (1/2)g t^2
g is about 9.81 m/s^2 so

h(t) = 73.5 + 9.8 t - 4.9 t^2

at ground, h(t) = 0
so
0 = 4.9 t^2 - 9.8 t - 73.5
and yes you can take 4.9 out approximately
0 = t^2 - 2 t -15
(t-5)(t+3) = 0
t = 5 seconds

Answer 2

factor out -4.9
-4.9(t^2+2t-15)
then think factors of -15 that will give you a positive 2. 5, -3 then set it up to zero because the height is at zero since it is on the ground.
-4.9(t+5)(t-3)=0
when solving you would get
t+5=0 which means t= -5 secs
t-3=0 which means t= 3 sec
Since time cannot be negative
your answer would be t= 3 secs

Answer 3

Meow