SOLUTION: I would really appreciate if someone would walk me through this problem in detail. I am stuck on this question. Thank you sooo much for your help!!!!! Near the surface of the ea

Question 985203: I would really appreciate if someone would walk me through this problem in detail. I am stuck on this question. Thank you sooo much for your help!!!!!
Near the surface of the earth, assuming negligible resistance from the air, the height in feet of a falling object is modeled well by the equation y = h − 16t^2 , where y is the height of the object, t is the number of seconds the object has been falling, and h is the height from which the object was dropped. (On your calculators, you will graph y for h and x for t).
a) If an iron ball were dropped from the Washington Monument, which is 555 feet high, how far above the ground would the ball be after 2 seconds of falling? How long would it take for the ball to hit the ground?
b) Due to air resistance, a falling bag of corn chips will not gain speed as rapidly as a falling iron ball. Cal Elayo, a student of science, found that the descent of a falling bag of chips is modeled well by the equation y = h − 2.5t^2 . In an historic experiment, Cal dropped a bag of chips from a point halfway up the Monument, while a friend simultaneously dropped the iron ball from the top. After how many seconds did the ball overtake the bag of chips?
c) Graph the equations y = 277.5 − 2.5t^2 and y = 555 −16t^2 on the same system of axes. Calculate the y- and t-intercepts of both curves. What is the meaning of these numbers? Notice that the curves intersect. What is the meaning of the intersection point?

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
y = h − 16t^2 , where y is the height of the object, t is the number of seconds the object has been falling, and h is the height from which the object was dropped
**********************************************************************************
a) y = 555 - 16(2^2)
y = 555 - 64 = 491,
--the iron ball is 491 feet above the ground after 2 seconds
0 = 555 - 16t^2
16t^2 = 555
t^2 = 555 / 16
t = square root(555 / 16) = square root(555) / 4 = 5.889609495 approx 5.9
--it takes 5.9 seconds for the iron ball to hit the ground
********************************************************************************
b) y = h − 2.5t^2, 555/2 = 277.5
set equations a and b equal to each other
555 - 16t^2 = 277.5 - 2.5t^2
13.5t^2 = 277.5
t^2 = 277.5 / 13.5 = 20.555555556
t = square root(20.555555556) = 4.533823503
--in 4.5 seconds the ball will overtake the bag of chips
*******************************************************************************
c) y = 555 - 16t^2 (red line), y = 277.5 - 2.5t^2 (green line)

****************************************************************
calculate y intercept by setting t = 0
calculate t intercept by setting y = 0
--y = 555 - 16t^2
y intercept is (555, 0)
--0 = 555 - 16t^2
t = 5.9
t intercept is (0, 5.9)
--y = 277.5 - 2.5t^2
y intercept is (277.5, 0)
--0 = 277.5 - 2.5t^2
t = 10.535653753 approx 10.5
t intercept is (0, 10.5)
****************************************************************
--y intercept is the height above the ground, t intercept is the time in seconds it takes the object to reach the ground
--the intersection point gives us the height above the ground when the objects over take each other and the time it takes for that to happen

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