Question

1. elmer is standing on the edge of a 30.0 m high cliff. he throws a rock vertically with an initial velocity of 12.0 m/s. a) how high does the rock go? b) how long does it take to reach its max height? 2. if you were to throw a large log over the edge of the grand canyon and it took 5.75 seconds to hit the ground, calculate the velocity of the log at impact in m/s and calculate the distance the log fell. 3. a tennis ball is dropped from a roof 16 meters from the ground. how long does it take for the ball to reach the ground?

Explanation:

Step1: Find height rock rises above cliff

Use the kinematic - equation $v^{2}=v_{0}^{2}-2gh$. At maximum height, $v = 0$. The initial velocity $v_{0}=12.0\ m/s$ and $g = 9.8\ m/s^{2}$.
$0=(12.0)^{2}-2\times9.8\times h_{1}$
$h_{1}=\frac{(12.0)^{2}}{2\times9.8}=\frac{144}{19.6}\approx7.35\ m$

Step2: Calculate total height of rock above ground

The cliff is $h_{0}=30.0\ m$ high.
$H=h_{0}+h_{1}=30.0 + 7.35=37.35\ m$

Step3: Find time to reach maximum height

Use the kinematic - equation $v = v_{0}-gt$. At maximum height, $v = 0$.
$0 = 12.0-9.8t$
$t=\frac{12.0}{9.8}\approx1.22\ s$

Step4: For log in Grand - Canyon

Use the kinematic - equation $v = v_{0}+gt$. Assuming the log is dropped ($v_{0}=0$) and $t = 5.75\ s$, $g = 9.8\ m/s^{2}$.
$v=0 + 9.8\times5.75=56.35\ m/s$
Use the kinematic - equation $y=v_{0}t+\frac{1}{2}gt^{2}$. Since $v_{0}=0$, $y=\frac{1}{2}\times9.8\times(5.75)^{2}= \frac{1}{2}\times9.8\times33.0625 = 161.90625\ m$

Step5: For tennis ball

Use the kinematic - equation $y=v_{0}t+\frac{1}{2}gt^{2}$. Since the ball is dropped ($v_{0}=0$) and $y = 16\ m$, $g = 9.8\ m/s^{2}$.
$16=\frac{1}{2}\times9.8\times t^{2}$
$t^{2}=\frac{16\times2}{9.8}=\frac{32}{9.8}$
$t=\sqrt{\frac{32}{9.8}}\approx1.81\ s$

Answer:

a) The rock goes $37.35\ m$ high above the ground.
b) It takes $1.22\ s$ to reach its maximum height.
For the log: The velocity at impact is $56.35\ m/s$ and the distance it fell is $161.90625\ m$.
For the tennis ball: It takes approximately $1.81\ s$ to reach the ground.