Question

a rock is dropped from the edge of a cliff that is 201 meters above water. use the position function below for free - falling objects, (s(t)=-4.9t^{2}+v_{0}t + s_{0}) (a) determine the position and velocity functions for the rock. (s(t)=) blank (v(t)=) blank (b) determine the average velocity (in m/sec) on the interval (5,8). blank m/sec (c) find the instantaneous velocities (in m/sec) when (t = 5) and (t = 8). (v(5)=) blank m/sec (v(8)=) blank m/sec (d) find the time (in sec) required for the rock to reach the surface of the water. (round your answer to two decimal places.) blank sec (e) find the velocity (in m/sec) of the rock at impact. (round your answer to two decimal places.) blank m/sec

Explanation:

Step1: Position function with initial conditions

$s(t) = -4.9t^2 + 201$ (since $v_0=0$, $s_0=201$)

Step2: Velocity function as derivative of position

$v(t) = s'(t) = -9.8t$

Step3: Average velocity on [5,8]

$\text{Average velocity} = \frac{s(8)-s(5)}{8-5}$. Calculate $s(8)=-4.9(8)^2+201=-112.6$, $s(5)=-4.9(5)^2+201=78.5$. Then $\frac{-112.6-78.5}{3}=-63.7$

Step4: Instantaneous velocity at t=5

$v(5) = -9.8(5) = -49$

Step5: Instantaneous velocity at t=8

$v(8) = -9.8(8) = -78.4$

Step6: Time to reach water (s(t)=0)

Solve $-4.9t^2 + 201 = 0 \Rightarrow t = \sqrt{\frac{201}{4.9}} \approx 6.41$

Step7: Velocity at impact

$v(t) = -9.8(6.41) \approx -62.77$

Answer:

s(t) = -4.9t² + 201
v(t) = -9.8t
-63.7
v(5) = -49 m/sec
v(8) = -78.4 m/sec
6.41
-62.77