I can not for the life of me wrap my head around this PLEASE help H = -16t + 368t

Question

One small piece of ash was ejected from the volcano with an initial velocity of 368 ft/sec. The height H, in feet, of our ash projectile is given by the equation:
H = -16t + 368t
Where t is the time in seconds. The graph of this equation will be a parabola. We will assume that the volcano has no height. In other words, when H=0 at t=0.
Question 1. When does the ash projectile reach its maximum height? t=? 
Question 2. What is its maximum height?
Question 3. When does the ash projectile return to the ground? 

Answer 1

Hi Wendy.

 

I think this needs some clarifying...   For this to be a parabola, the formula should really be   

 

h = -16t² + 368t

 

 

Since this is a parabola, it is a parabola that points down (looks like a hill, or an upside-down "U"), there is a maximum height at the vertex.  And the formula for the x-coordinate of the vertex of a parabola is given by

x = -(b/2a),  but in this case, we are using "t" instead of "x", a is (-16) and b is 368

 

QUESTION 1:  So the maximum height occurs at  t = -(368/2(-16))  = 11.5 seconds

 

If this is a calculus class, you would find dh/dt = -32t + 368 and set it = 0, and get the same thing.

 

 

Now that you know how long it takes to reach the maximum height, plug the "t" you got in QUESTION 1 into the equation for h.

 

QUESTION 2:   h = -16(11.5)² +368(11.5) =2116 feet.

 

 

When the ash projectile returns to the ground...  What height is the ground?

 

Well, it is at h = 0,  so plug  h = 0 into the equation given for h, and see what you get.

 

0 = -16t² +368t

 

0 = t(-16t+368)

 

So it is on the ground at t = 0 (that is when it started)  or

when  -16t+368=0, which solves to be  t = 23 seconds.

 

QUESTION 3:  Returns to ground at t=23 seconds.