## anonymous 4 years ago A projectile is thrown upward so that its distance above the ground after t seconds is h = - 16t2 + 440t. After how many seconds does it reach its maximum height?

1. anonymous

few ways to solve this

2. anonymous

are you familiar with derivatives?

3. anonymous

with a graph correct?

4. anonymous

Without Calculus y|dw:1349657074734:dw|ou know that this graph will look like this

5. anonymous

oh ok

6. anonymous

so you know it hits its maximum height at its' vertex. you know how to find the x for the vertex by $x=\frac{-b}{2a}$

7. anonymous

x so happens to be time in our case =P

8. anonymous

ohhhh

9. anonymous

|dw:1349657191197:dw|

10. anonymous

21 seconds

11. anonymous

with calculus $y=-16t^2+440t$ $y'=-32t+440=velocity$ we know that the velocity of y is 0 at the point of max height so $0=-32t+440$ $-440=-32t$ $\frac{-440}{-32}=t$ 13.75 seconds

12. anonymous

without calculus $ax^2+bx+c$ in our case $ax^2+bx$ $x=\frac{-b}{2a}$ $\frac{-440}{-32}$ 13.75 seconds

13. anonymous

oh wow i was completly of thanks alot !!!