SOLUTION: the function S(x)=-4.9t^2+4.9t+58.8 gives the height S, in meters of an object launched with velocity of 4.9 meters per second from a height of 58.8 meters.
A)Hw long will it tak
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-> SOLUTION: the function S(x)=-4.9t^2+4.9t+58.8 gives the height S, in meters of an object launched with velocity of 4.9 meters per second from a height of 58.8 meters.
A)Hw long will it tak
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Question 1138638: the function S(x)=-4.9t^2+4.9t+58.8 gives the height S, in meters of an object launched with velocity of 4.9 meters per second from a height of 58.8 meters.
A)Hw long will it take the object to hit the ground?
B)find the interval on which the height of the object is greater than 49m.
please show your work Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! the function S(x)=-4.9t^2+4.9t+58.8 gives the height S, in meters of an object launched with velocity of 4.9 meters per second from a height of 58.8 meters.
A)Hw long will it take the object to hit the ground?
B)find the interval on which the height of the object is greater than 49m.
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S(x)=-4.9t^2+4.9t+58.8 has no x terms on the right side.
Use S(t)=-4.9t^2+4.9t+58.8 with t in seconds.
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A)How long will it take the object to hit the ground?
S(t)=-4.9t^2+4.9t+58.8 = 0 at impact.
Solve for t
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B)find the interval on which the height of the object is greater than 49m.
S(t)=-4.9t^2+4.9t+58.8 = 49
Solve for t, call it T seconds.
0 < t < T
