Question 266400: An object's altitude, in meters, is given by the polynomial h + vt - 9.8t^2, where h is the height in meters from which the launch occurs, v is the initial upward speed in meters per second, and t is the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 175 meters tall. If the initial speed is 32 meters per second, how high above the ground will the pebble be after 3 seconds? Round results to the nearest tenth of a meter.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! An object's altitude, in meters, is given by the polynomial h = vt - 9.8t^2, where h is the height in meters from which the launch occurs, v is the initial upward speed in meters per second, and t is the number of seconds for which the rocket is airborne.
A pebble is shot upward from the top of a building 175 meters tall. If the initial speed is 32 meters per second, how high above the ground will the pebble be after 3 seconds? Round results to the nearest tenth of a meter.
:
The equation we want to find the height above the ground:
h = -9.8t^2 + 32t + 175
Replace t with 3 sec
h = -9.8(3^2) + 32(3) + 175
h = -88.2 + 96 + 175
h = 182.8 ~ 183 meters above the ground
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