Jamwright13
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One small piece of ash was ejected from the volcano with an initial velocity of 336ft/sec. The height H, in feet, of the ash projectile is given by the equation: H= -16t^2 + 336t Where t is the time in sec. The graph of this equation will be a parabola. We will assume that the volcano has no height. H= 0 at t= 0.
1. When does the ash projectile reach its maximum height?
2.What is its maximum height?
3.When does the ash projectile return to the ground?

Respuesta :

Answer: I only know number one t=10.5 seconds

Answer:

(a)t=10.5 seconds

(b)1764 feet

(c)t=21 seconds

Step-by-step explanation:

The equation of the projectile is given by:

[tex]H= -16t^2 + 336t[/tex]

(1)The ash projectile reaches its maximum height when the slope of the line is equal to zero.

First, we find the derivative of H.

[tex]H^{'}= -32t + 336[/tex]

When [tex]H^{'}=0[/tex]

[tex]= -32t + 336=0\\-32t=-336\\[/tex]

Divide both sides by -32

t= 10.5 seconds

(2) The maximum height occurs at the point where t= 10.5

[tex]H= -16t^2 + 336t\\H= -16(10.5)^2 + 336(10.5)\\Max(H)=1764 feet[/tex]

(3)The ash projectile returns back to the ground when its Height, H =0

[tex]H= -16t^2 + 336t\\\\-16t^2 + 336t=0\\-16t^2 =- 336t\\-16t=-336\\t=21 seconds[/tex]

The ash projectile returns to the ground after 21 seconds.

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