looking over the edge of a cliff, you become curious as to how high up you are. you take a rock and drop it over the edge and count. The rock strikes the ground in between 3 and 4 of your count. so let's estimate that it took 3.5s for it to hit the ground. using the bottom of the cliff as y= 0m, estimate the height of the cliff assuming the rock was in freefall. start by identifying each known kinematic value from when the rock was falling:
a. is y0 known if so what is it?
b. is yf known if so what is it? 0m
c. is v0 known if so what is it?
d. is vf known if so what is it?
e is a known if so what is it? -9.8m/s
f. is t known if so what is it? 3.5s
This is as far as I got with my answers how can I figure the rest?
yf = yo + vo t + (1/2) a t^2
yf = 0
yo = unknown
vo = 0 (dropped not thrown)
vf irrelevant ( = a t)
a = -4.9 indeed
t = 3.5
so
0 = yo + 0 -4.9 (3.5)^2
or
yo = 4.9 (3.5)^2
To estimate the height of the cliff, we can use the kinematic equation for vertical motion:
y = y0 + v0t + (1/2)at^2
Let's assign the known values:
a. y0: The initial position of the rock. Since we are taking the bottom of the cliff as y = 0m, the initial position would be the height of the cliff. However, the height of the cliff is not given, so y0 is unknown for now.
b. yf: The final position of the rock when it strikes the ground. We know it is 0m, as stated in the problem.
c. v0: The initial velocity of the rock. When the rock is dropped, its initial velocity is 0m/s because it starts from rest.
d. vf: The final velocity of the rock when it strikes the ground. We don't have this information yet.
e. a: The acceleration due to gravity is a known value, -9.8m/s^2. It is always negative because it acts downward.
f. t: The time taken by the rock to hit the ground is 3.5s, as given in the problem.
We need to find y0 (the height of the cliff) and vf (the final velocity of the rock when it strikes the ground). To do this, we can use two equations of motion:
1. vf = v0 + at
2. y = y0 + v0t + (1/2)at^2
Let's start with the first equation:
vf = v0 + at
Substituting the known values:
0m/s (vf) = 0m/s + (-9.8m/s^2)(3.5s)
Now we can solve for v0:
0 = v0 - 34.3
v0 = 34.3m/s
Now that we have the value of v0, we can plug it into the second equation:
y = y0 + v0t + (1/2)at^2
0m = y0 + (34.3m/s)(3.5s) + (1/2)(-9.8m/s^2)(3.5s)^2
Simplifying the equation:
0 = y0 + 119.05m - 60.53m
Rearranging the equation:
y0 = -119.05m + 60.53m
y0 ≈ -58.52m
Since the height of the cliff cannot be negative, we discard the negative sign. Therefore, the estimated height of the cliff is approximately 58.52 meters.