Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration. So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.
Answer:
Heat is transferred by the hot air or water moving to a cooler area. The elements rotate in circular motions, giving the geyser pressure.
Answer:
a=F/m
a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)
a=4m/s^2
Explanation: