Question

Eyjafjallajokull is a volcano in Iceland. In a recent eruption, a projectile is ejected with an initial velocity of 304 feet per second. The height, H, in feet is given by the equation: H=-16t^2+304t Where t is the time in seconds. We will use the equation that traces the path of the projective to determine the following information: 1) The time it takes the projective to reach its highest point. (t=?) 2) The maximum height? (H=?) 3) The time it takes the projectile to return to the ground? (t=?)

Answer 1

Answer:

A volcano ejected with an initial velocity of 304 feet per second.

The height in feet is given by the equation H=-16t^2+ 304t where t is the time in seconds.

We will use the equation that traces the path of the projectile to determine the following information:

1) The time it takes the projectile to reach its maximum height (using the vertex formula: t=-b/2a)

The equation; H = -16t^2 + 304t, a=-16; b=304

t = -304/2°(-16)

t = 9.5 sec to reach max height

2) The maximum height of the projectile (use the result from part 1 to find the maximum height of the projectile)

t = 9.5

H = -16(9.5^2) + 304(9.5)

h = = -1444 + 2888

h = 1444 ft is the max height

3) Find the time it takes the projectile to return to the ground. Assume that the projectile starts at height H=0.

-16t^2 + 304t = 0

factor out -16t

-16t(t - 19) = 0

t = 19 sec to reach the ground

Step-by-step explanation: