a volcano erupts, one piece of ash was ejected from the volcano with an initial velocity 368ft per/sec... The height is given with the equation H=-16t^2 +368t

Question

We need to find out when does the ash projectile reach its maximum height. What is the maximum height. Last, when does the ash projectile reach the ground.

 

H=height in feet 

T= time in sec

This is a Parabola

Answer 1

So, we start by finding the x-value of the vertex.

 

x = -b/2a

x = -368/(2(-16) = 368/32 = 11.5

 

What's the height (y-value) after 11.5 seconds?

 

H = -16(11.5²) + 368(11.5)

H = -16(132.25) + 4232

H = -2116 + 4232

H = 2116

 

When it reaches the ground, the height is 0 (duh!).

 

0 = -16t² + 368t

0 = -16t(t - 23)

 

-16t = 0

t = 0

Yep, we know that.

 

t - 23 = 0

t = 23 seconds