Question 414098: An objects altitude, in meters is given by the polynomial h+vt-9.8t, where is the height in meters from which the launch occurs, v is the initial upward speed in meters per second, and the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 183 meters tall. If the initial speed is 39 meters per second, how high above the ground will the pebble be after 3 seconds?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! An objects altitude, in meters is given by the polynomial h+vt-9.8t, where is the height in meters from which the launch occurs, v is the initial upward speed in meters per second, and the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 183 meters tall. If the initial speed is 39 meters per second, how high above the ground will the pebble be after 3 seconds?
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I don't know where you got that equation, but:
h(t) = h+vt-9.8t
h(3) = 183 + 39*3 - 9.8*3
= 270.6 meters
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I would use h(t) = h + vt - 4.9t^2
h(3) = 183 + 39*3 - 4.9*9
= 255.9 meters
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