## babydoll332 3 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. Outkast3r09

few ways to solve this

2. completeidiot

are you familiar with derivatives?

3. babydoll332

with a graph correct?

4. Outkast3r09

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

5. babydoll332

oh ok

6. Outkast3r09

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. Outkast3r09

x so happens to be time in our case =P

8. babydoll332

ohhhh

9. Outkast3r09

|dw:1349657191197:dw|

10. babydoll332

21 seconds

11. Outkast3r09

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. Outkast3r09

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

13. babydoll332

oh wow i was completly of thanks alot !!!