SOLUTION: the height in feet of a projectile after t seconds is given by h(t)=-16t^2 +96t determine the values for which the projectile is at the ground. (the -16 has an exponent of 2 I

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Question 113289This question is from textbook Beginning and Intermediate Algebra
: the height in feet of a projectile after t seconds is given by h(t)=-16t^2 +96t
determine the values for which the projectile is at the ground.
(the -16 has an exponent of 2 I couldn't get it to show up right also I need to set this problem up as an equation the chapter we are working on has to do with factoring.) This question is from textbook Beginning and Intermediate Algebra

Found 2 solutions by SHUgrad05, Earlsdon:
Answer by SHUgrad05(58) About Me  (Show Source):
You can put this solution on YOUR website!
The projectile would be at the ground at h(t)=0.
-16t^2+96^t=0
First factor out 16t
16t(-t+6)=0
16t=0 and -t+6=0
the projectile is at the ground when
t=0 and t=6

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The height of an object, as a function of time, t, propelled upwards from an initial height h%5B0%5D with an initial velocity v%5B0%5D is given by:
h%28t%29+=-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
In your problem:
v%5B0%5D+=+96ft.
h%5B0%5D+=+0ft.... or on the ground.
You want to find the values of t for which h = 0, so substitute h=0 and solve for t.
0+=+-16t%5E2%2B96t Factor a t.
0+=+t%28-16t%2B96%29 Apply the zero product principle.
t+=+0 and
16t+=+96 Divide both sides by 16.
t+=+6
So the height of the projectile is 0 (on the ground) at times, t = 0 secs and t = 6 secs.