Question

ii. lesson 7: hand - in assignment

1. a rock is thrown downward from a cliff at 15.0 m/s. the rock hits the waves below after 2.45 s.

a) what was the height of the cliff?
b) what was its velocity after 2.0 s?

Explanation:

Step1: Identify the kinematic - equation for part a

We use the equation $y = v_0t+\frac{1}{2}gt^2$, where $v_0$ is the initial velocity, $t$ is the time, and $g = 9.8\ m/s^2$ (acceleration due to gravity). Here, $v_0=15.0\ m/s$, $t = 2.45\ s$, and $g = 9.8\ m/s^2$.
$y=(15.0\times2.45)+\frac{1}{2}\times9.8\times(2.45)^2$

Step2: Calculate the value of $y$ for part a

$y = 15.0\times2.45+4.9\times6.0025$
$y=36.75 + 29.41225$
$y = 66.16225\approx66.2\ m$

Step3: Identify the kinematic - equation for part b

We use the equation $v = v_0+gt$. Here, $v_0 = 15.0\ m/s$, $t = 2.0\ s$, and $g = 9.8\ m/s^2$.
$v=15.0+9.8\times2.0$

Step4: Calculate the value of $v$ for part b

$v=15.0 + 19.6$
$v = 34.6\ m/s$

Answer:

a) $66.2\ m$
b) $34.6\ m/s$