Question

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?

Answer 1

Answer: I only know number one t=10.5 seconds

Answer 2

(a)t=10.5 seconds

(b)1764 feet

(c)t=21 seconds

Explanation:

The equation of the projectile is given by:

`H= -16t^2 + 336t`

(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.

`H^(')= -32t + 336`

When
`H^(')=0`

`= -32t + 336=0\\-32t=-336\\`

Divide both sides by -32

t= 10.5 seconds

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

`H= -16t^2 + 336t\\H= -16(10.5)^2 + 336(10.5)\\Max(H)=1764 feet`

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

`H= -16t^2 + 336t\\\\-16t^2 + 336t=0\\-16t^2 =- 336t\\-16t=-336\\t=21 seconds`

The ash projectile returns to the ground after 21 seconds.