Question

A cannon launches a cannonball from a height of 3 feet with an upward velocity of 40 feet per second. The function h=-16t^2+40t+3 gives the height "h" (in feet) of the cannonball after "t" seconds.
a. After how many seconds is the cannonball 10 feet above the ground?

b. What is the maximum height of the cannonball?

Please answer with the correct unit!

Answer 1

Using the voluma given h=-16t²+40t+3, we can find the seconds after the h=10 by subsituting 10 as h, then isolating t

h=-16t²+40t+3
10=-16t²+40t+3             h=10 was inserted
-16t²+40t+3-10=0         10 was moved to the right side
-16t²+40t-7                   Like terms were collected
-(16t²-40t+7)                 Factored out -1 (completely optional)

Then complete the square
-(16t²-40t)-7                  Moved -7
-4(2t²-10t)-7                  Factored out 4
-8(t²-5t)-7                      Factored out 2
-8(t²-5t+$\frac{25}{4}$)-7 +$\frac{25}{4}$
-8(t-5/2)² -7+ $\frac{25}{4}$
-8(t-5/2)² -7+ $6\frac{1}{4}$
-8(t-5/2)²-$\frac{3}{4}$

 OR 
You could use the QUADRATIC FORMULA
(Legit so fun....)    
t=$\frac{-b \frac{+}{}\sqrt{b^2-4ac} }{a^2}$  (a=16, b=-40, c=7)
t=$\frac{-(-40) \frac{+}{}\sqrt{-40^2-4(16)(7)} }{2(16)}$ 
t= $\frac{40\frac{+}{}\sqrt{1600-448} }{32}$  
t= $\frac{40\frac{+}{}\sqrt{1152} }{32}$  

1) Time is therefore $\frac{40+\sqrt{1152} }{32}$  and $\frac{40-\sqrt{1152} }{32}$ 


When finding the maxumium height, there is a rule that when h$_{max}$ is achieved the velocity is ALWAYS ZERO. Therefore when graphing the path of the ball the vertex's y-coordinate will be the maximum height, and the x-coordinate is the value of time at that particular height. 

2) The maximum height is 28 meters reached at 1.25 seconds

Answer 2

A)  We want to know when h=10 so

10=-16t^2+40t+3

16t^2-40t+7=0

Using the quadratic formula:

t=(40±√1152)/32 

t≈0.189 and 2.311 seconds

So it is at 10 ft twice, about 2/10 of a second when it is rising and at 2 3/10 of a second when it is falling...

b.

The maximum height of the ball is when velocity is equal to zero, or dh/dt=0

dh/dt=-32t+40, dh/dt=0 when 32t=40, t=40/32=5/4=1.25 seconds

h(1.25)=28ft

So the maximum height of the ball occurs after 1.25 seconds and reaches a height of 28 feet.