Question

The height, h, in meters above the ground, of a projectile at any time, t, in seconds, after launch is defined by the function h = -4t^2+ 48t+ 3.l Find the maximum height reached by the projectile.

Answer 1

do you know calculus...?

Answer 2

agh i need help with some explanation, as I'm doing this course online and the tutorials aren't helping much :/

Answer 3

Then watch this prof :
http://www.youtube.com/watch?v=jbIQW0gkgxo
he will give the concept very well

Answer 4

Completing the square will give you the maximum height (at the vertex):
\[h(t)=-4t^2+48t+3=-4(t-6)^2+147=-4(t-h)^2+k\]
where (h,k) is the vertex, or where the projectile reaches its maximum.
=> (h,k)=(6,147)
The projectile reaches 147 m, same result if calculus was used.
If completing the square appears difficult, you can always take the short cut in that the value of t at the vertex = -b/2a = -48/(2*-4) = 6, and find h(6) accordingly.

Answer 5

ahhh thanks for your help! :D

Answer 6

i really appreciate it xD