Question

A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff. A second rock is thrown vertically from the cliff, and it takes this rock 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock thrown, assuming no air resistance?

A) 12.3 m/s upward
B) 4.76 m/s downward
C) 4.76 m/s upward
D) 12.3 m/s downward
E) 5.51 m/s downward

Answer 1

D) 12.3 m/s downward

Step-by-step explanation:

We use the next free fall equation

`h=v_(0)t+(1)/(2)gt^2`

where
`h` is the height of the cliff,
`v_(0)` the initial velocity,
`g` the acceleration of gravity (
`g=9.81m/s^2`) and
`t` is time.

For the fist rock
`v_(0)=0` since the rock was dropped, and
`t=3s`, so we have:

`h=(0m/s)(3s)+(1)/(2)(9.81m/s^2)(3s)^2`

simplifying

`h=44.145m`

the height of the cliff is 44.145m

Now, about the second rock we know that is the same height and now the time es
`t=2s`, and we need to find
`v_(0)`

From the fist equation

`h=v_(0)t+(1)/(2)gt^2`

we clear for
`v_(0)`

`v_(0)t=h-(1)/(2) gt^2\\v_(0)=(h)/(t) -(1)/(2) gt`

and substitute known values

`v_(0)=(44.145m)/(2s) -(1)/(2) (9.81m/s^2)(2s)`

`v_(0)=22.07m/s -9.81m/s`

`v_(0)=12.26m/s` wich rounds up to
`v_(0)=12.3m/s`

the direction is downward because the rock is thrown so that it falls through the cliff.