50 POINTS QUICK PLEASE HURRY A rocket is launched vertically from the ground with an initial velocity of 64.

(a) Write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched.

(b) Graph on the coordinate plane.

(c) Use your graph from Part 3(b) to determine the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air.

We use the formula;

s = ut + 1/2at²

s = distance

u = initial velocity

t= time

a = acceleration (in case we use acceleration due to gravity = 10m/s²)

since the rocket is launched upward, it is decelerating, so the equation becomes;

s = ut - 1/2at²

take s = height = y

t= time = x

64ft/sec = 19.5m/s

the equation becomes;

y = 19.5x - 1/2(10)x²

or

y = -5x² + 19t

the graph is attached;

from the graph

c) maximum height = 19m = 62.3ft

time taken to reach max. height = 1.95 sec

time the rocket was in air = 3.9 sec.

a rocket is launched vertically from the ground with an initial velocity of 64 ft/sec .

Step-by-step explanation:

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step-by-step explanation:

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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