An object launched straight puwar from the ground level wiht an initional velocity of 32 feet per second.The formula H=32T-16T^2 gives it heigtht,h,in feet after t second. At what time does the object hit the ground?
Question
Leave a Comment
You must be logged in to post a comment.
Answer:
2 secs after
Step-by-step explanation:
Given the height of the object modeled by the eqaution;
H(t) = 32T – 16T² where;
T is the time in seconds
The object hits the ground when h(t)= 0
Substitute H = 0 into the equation and get t as shown;
0 = 32T – 16T²
0-32T = -16T²
-32T = -16T²
32T = 16T²
32 = 16T
T = 32/16
T = 2secs
Hence the object hits the ground after 2 secs