SOLUTION: An airplane is flying at an altitude of 1000 meters. A skydiver jumps from the plane with no upward speed. The equation h(x)=-4.9x^2 + 1000 where h(x) represents her height above t

Question 1071607: An airplane is flying at an altitude of 1000 meters. A skydiver jumps from the plane with no upward speed. The equation h(x)=-4.9x^2 + 1000 where h(x) represents her height above the ground during the freefall as a function of time, x, in seconds, since the beginning of her jump.
5. How high above the earth is the skydiver after 6.5 seconds?
6. After how many seconds is the skydiver 100 meters above the ground?
7. After how many seconds does the skydiver land?
please get back to me soon. thanks!
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
5. After 6.5 seconds, h(6.5)=-4.9(6.5^2)+1000=-207.025+1000=793 meters to nearest meter.
6. 100=-4.9 x^2+1000
-900=-4.9 x^2
x^2=183.67
x=13.55 sec.
7. Assuming that there is no change in speed, no deceleration,
0=-4.9x^2+1000
4.9x^2=1000
x^2=204.08
x=14.29 sec
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