Question

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is 300 th all How much earker does the thrown rock strike the ground?

Answer 1

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

• $y_{i}=300m$

• $y_{f}=0m$
• $v_{iD}=0m/s$
• $v_{iT}=-29m/s$, it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$, to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$, then clearing for $t_{D}$.

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$, to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$, then, $0=-4.9t_{T}^{2}-29t_{T}+300$, as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$