Answer:
The thrown rock will strike the ground [tex]2.42s[/tex] earlier than the dropped rock.
Explanation:
Known Data
Time of the dropped Rock
We can use [tex]y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}[/tex], to find the total time of fall, so [tex]0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}[/tex], then clearing for [tex]t_{D}[/tex].
[tex]t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s[/tex]
Time of the Thrown Rock
We can use [tex]y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}[/tex], to find the total time of fall, so [tex]0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}[/tex], then, [tex]0=-4.9t_{T}^{2}-29t_{T}+300[/tex], as it is a second-grade polynomial, we find that its positive root is [tex]t_{T}=5.4s[/tex]
Finally, we can find how much earlier does the thrown rock strike the ground, so [tex]t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s[/tex]
