SOLUTION: After opening the parachute, the descent of a parachutist follows a linear model. At 2:08 PM the height of the parachutist is 7000 feet. At 2:10 PM the height is 4600 feet. a) Wri

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Question 433715: After opening the parachute, the descent of a parachutist follows a linear model. At 2:08 PM the height of the parachutist is 7000 feet. At 2:10 PM the height is 4600 feet.
a) Write a linear equation that gives the height of the parachutist in terms of time t. (Let t = 0 represent 2:08 PM and let t be measured in seconds.)
b) Use the equation in part (a) to find the time when the parachutist will reach the ground.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You are given 2 points on the line:
(0,70.00) and (120,46.00)
Height is in hundreds of feet, and time is in seconds
The formula is:
+%28+h+-+46+%29+%2F+%28+t+-+120+%29+=+%28+70+-+46+%29+%2F+%28+0+-+120+%29+
+%28+h+-+46+%29+%2F+%28+t+-+120+%29+=+24+%2F+%28-120%29+
+-120%2A%28+h+-+46+%29+=+24%2A%28t+-+120+%29+
+5520+-+120h+=+24t+-+2880+
+120h+=+-24t+%2B+8400+
+5h+=+-t+%2B+350+
+h+=+%28-1%2F5%29%2At+%2B+70+ answer ( h is in hundreds of feet )
check:
(120,46)
+46+=+%28-1%2F5%29%2A120+%2B+70+
+46+=+70+-+24+
+46+=+46+
OK
When does +h+=+0+
+0+=+%28-1%2F5%29%2At+%2B+70+
+%281%2F5%29%2At+=+70+
+t+=+350+ seconds
or, 5 minutes and 50 seconds