A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. an echo is heard 2.80 s later. how far away is the ref
1 answer:
The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
The speed of the sound wave is:
, and since it is moving by uniform motion, we can find the distance covered by the wave using
And we said this corresponds to twice the distance between us and the reflecting object, so:
so, the object is 480 meters away.
The speed of the sound wave is:
And we said this corresponds to twice the distance between us and the reflecting object, so:
so, the object is 480 meters away.
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Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
We must consider that it's launched from the ground () and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 () and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get , we do:
And we substitute the values: