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:
D. scalene
e. isosceles
f. isosceles
g. equilateral
h. scalene
i. equilateral
40 i think because when you divide 400 by 10 it equals 40
Angle q and angle t
angle j and angle k
