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

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

Step-by-step 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-(gt^(2))/(2)`, to find the total time of fall, so
`0=300m-((9.8m/s^(2))t_(D)^(2))/(2)`, then clearing for
`t_(D)`.

`t_(D)=\sqrt[2]{(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-(gt^(2))/(2)`, to find the total time of fall, so
`0=300-29t_(T)-((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`