Help! The distance an object is from the ground as it falls toward the ground can be found using this formula.

Question

d= -16t^2+100t+6 where t is measured in secondsand d is measured in feet. Find the distance traveled when t=0 seconds, then 1 sec, 2 sec, 3 sec, 5 sec, and 6 sec. Explain what your results mean in terms of this problem! I especially need help explaining the answers. What is the difference between and what does it mean when you get a positive number for an answer as opposed to a negative number? Please Help!!

Answer 1

What this equation is describing is this:

An object is launched into the air at time T=0.
The object is launched from a height of 6 feet.
The object has a launch velocity of 100 feet/second vertically.
(The equation does not tell us of any horizontal component, but the vertical component is 100ft/s.)
Plugging in different values for T results in different values for the altitude (example: after 1 second the object has an altitude of 90 feet).
If you plug in a number higher than about 6.31, the results become negative.
This means that the object has come back down to ground level (and would continue to go down if the ground were not in the way).

Answer 2

y = 1/2 a t^2 +v0*t +y0

a = acceleration
t = time
v0 = initial speed
y0 = initial height