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

Answer:

D) 12.3 m/s downward

Explanation:

We use the next free fall equation

$h=v_{0}t+\frac{1}{2}gt^2$

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:

$h=(0m/s)(3s)+\frac{1}{2}(9.81m/s^2)(3s)^2$

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

$h=v_{0}t+\frac{1}{2}gt^2$

we clear for $v_{0}$

$v_{0}t=h-\frac{1}{2} gt^2\\v_{0}=\frac{h}{t} -\frac{1}{2} gt$

and substitute known values

$v_{0}=\frac{44.145m}{2s} -\frac{1}{2} (9.81m/s^2)(2s)$

$v_{0}=22.07m/s -9.81m/s$

$v_{0}=12.26m/s$ 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.