## anonymous one year ago A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 9. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball's maximum height? A) 1 s; 22 ft B) 2 s; 22 ft C) 2 s; 6 ft D) 1 s; 54 ft

1. arindameducationusc

is this a physics question?

2. Michele_Laino

from your equation of motion, I see that the initial speed is 36 feet/sec

3. Michele_Laino

now, the equation for speed at a generic time t, is: $\Large v\left( t \right) = {v_0} - gt \to v\left( t \right) = 36 - 32t$

4. Michele_Laino

in order to compute the time t, at which the ball has reached its maximum height, you have to set v=0, so you have to solve this equation: $\Large 0 = 36 - 32t$ please solve for t

5. anonymous

No, Algebra @arindameducationusc

6. anonymous

How am I supposed to do that.. can u help me please @Michele_Laino

7. Michele_Laino

the general formulas of kinematics, are: $\Large \begin{gathered} h\left( t \right) = - \frac{1}{2}g{t^2} + {v_0}t + {h_0} \hfill \\ v\left( t \right) = {v_0} - gt \hfill \\ \end{gathered}$

8. Michele_Laino

now, when the ball reaches its maximum height, its speed is zero, right?

9. Michele_Laino

from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec

10. anonymous

Could there not also be a horizontal component of initial velocity such that the vertical component is 32 m/s yet and horizontal component is such that initial velocity is 36 m/s?

11. anonymous

From a mathematical perspective, a) is asking for the t-coordinate of the vertex. This is given by $t=-\frac{ b }{ 2a }$with b=36 and a=-16 from the given function.

12. IrishBoy123

13. IrishBoy123

if this is algebra, you must be doing something like completing the square?? pls advise.

14. anonymous

15. IrishBoy123

so complete the square for $$h = –16t^2 + 36t + 9$$

16. IrishBoy123

helps to take the leading coefficient outside, as in $$\large h(t)=–16(t^2-\frac{36}{16}t-\frac{9}{16})$$ lots of square numbers in there so should be fine...

17. anonymous

oh my

18. anonymous

my guess is that this has nothing to do with calculus it is really not this complicated $h = –16t^2 + 36t + 9$ is a parabola that opens down the maximum is the second coordinate of the vertex the first coordinate of the vertex of $y=ax^2+bx+c$ is always $-\frac{b}{2a}$ which, in our case is $-\frac{36}{2\times (-16)}=1$

19. anonymous

ok not 1 sorry

20. anonymous

$\frac{36}{32}=1.25$

21. anonymous

that answers the first question a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

22. anonymous

to find the maximum height replace $$t$$ by $$1.25$$ i see your answer choices do not have $$1.25$$ so either there is a mistake in the answers, or it should have been $h=-16t^2+32t+9$