Stone is thrown vertically upward with speed = 11.5 m/s from edge of cliff 67.0 m high. Total distance travel?

Question

I already got the first 2 questions:
(a) How much later does it reach the bottom of the cliff?
i got 5.053 s
(b) What is its speed just before hitting?
38.1094

I just don't know how ro find the tital distance. Is it find the dinstance and then add 67.0 m ? I don't understand so plaese help me out i would really appreciate it Thanks :) xoxo

Answer 1

newtons laws of motion problem

time to apex
vf = 0
vi = 11.5 m/s
g = 9.81

vf = vi - gt
t = vi/g = 11.5/9.81 = 1.17

distance up
s = vi*t - 1/2gt^2
s = 11.5*1.17 - 1/2*9.81*1.17^2 = 6.74 m

height above the bottom of the cliff
s = 6.74+67.0 = 73.74 m

time to the bottom,
since s is downward we can treat it negatively
s = vi*t - 1/2gt^2
t = sqrt(2s/g) = sqrt(2*73.74/9.81) = 3.88 s

velocity at the bottom

vf = vi - g*t
vf = -9.81*3.88 = -38.0 m/s (negative implies downward direction)

a) total time traveled = 1.17+3.88 = 5.05

b) speed = 38.0 m/s

c) total distance traveled = 73.74 + 6.74 = 80.48 m

Answer 2

Your first two answers are correct and for the third all you need to do is find out how far above the cliff the stone changed directions. To do that I would use the equation:

V^2 = Vo^2 + 2*a*deltaX

Since the stone is changing directions at this point the V will be 0

Plug in 11.5 for Vo and -9.8 for a

This will give you a delta X of about 6.75 meters. So the stone went 6.75 meters above the cliff, then 6.75 meters down to the top of the cliff, then 67 meters to the bottom of the cliff. This gives a total distance traveled of 80.5 meters

Answer 3

Yes. If the stone travels "s" meters upwards then it will have to travel "s" meters downwards to be level with the top of the cliff again. It will still have to fall another 67 meters to reach the bottom of the cliff. So the total distance traveled is 2s + 67 meters.

I/m sure you already know all this since you worked out the other parts correctly.

v0 = gt so t = v0/g

s = v0t - (1/2)gt^2 = v0^2/g - (1/2)g(v0/g)^2
2sg = 2v0^2 - v0^2
s = v0^2/2g = (11.5)^2/(2*9.8) = 6.748 meters

Total distance = 2(6.75) + 67 = 80.495 meters

Answer 4

Kinetic energy (max) = Potential energy (min)
use 1/2mv^2 = mgh

h = 1/2 x 11.5^2
This would give you the distance traveled when the stone is thrown upwards

Total distance traveled
= h + 67.0
= 133.125 m

PS: I did not check the answer for part a and b.

Answer 5

No. From the observer's attitude it may nonetheless take 2 years for the tourist to return hence the tourist isn't vacationing quicker than the fee of sunshine in spite of the certainty that the tourist is ageing at a slower fee. The tourist could desire to think of it took decrease than 2 years as a results of fact of time dilation, however the certainty keeps to be is that it nonetheless took 2 years to make the around holiday.

Answer 6

They probably want you to write the answer as

D = √(v²/2g) + [ √(v²/2g) + h] = √(2v²/g) + h