help in physics?

Question

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70m high.
A) How much later does it reach the bottom of the cliff?
B) What is its speed just before hitting?
C) What total distance dit it travel?
Please show work

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70m high.
A) How much later does it reach the bottom of the cliff?
B) What is its speed just before hitting?
C) What total distance dit it travel?
Please show work

Ignore air resistance

Answer 1

Start with your basic equations:

y = y0 + v0 * t - (1/2)* g* t^2

y0 = the height you start at
v0 = the velocity you start at
g = 9.8 m.sec^2 , the acceleration due to gravity

y and t are the variables

Start with A)

The bottom of the cliff is when y = 0, so solve for "t":

0 = 70 + 12t - 4.9t^2

t = [ -12 +/- sqrt( (12^2) - (4)(-4.9)(70) ) ] / (2)(-4.9)
t = [ -12 +/- sqrt( 144 + 1372 ] / -9.8
t = [ -12 +/- sqrt( 144 + 1372) ] / -9.8
t = [ -12 +/- 38.94 ] / -9.8
t = 5.20 seconds or -2.74 seconds

Since time cannot be negative here, t = 5.20 seconds.

B)
Velocity is the derivative of the position function:

y' = v = 12 - 9.8t
at t = 5.20 seconds, v = -38.96, the negative sign indicates the direction of the ball is toward the ground, but the speed is just 38.96 m/s

C)
The distance has two parts:

The distance traveled before it hit its max height and the distance after it hit the max height.

Again, the max height is found by the derivative of the position function, and setting it equal to zero (because at max height, speed is also zero)
y' = v = 12 - 9.8t = 0
9.8t = 12
t = 1.22seconds

so y(1.22) = 70 + 12(1.22) - 4.9(1.22)^2
=77.35 meters

So it travels 77.35 - 70 = 7.35 meters on its way up.
and 77.35 meters on the way down:

Total distance = 77.35 + 7.35 = 84.7 meters.

Answer 2

Apparently the guy above me didn't learn much in school. This is an assumed basic physics problem. No drag on the rock so terminal velocity need not be taken into account. The weight would only matter if drag was taken into account, and it isnt so weight is not an issue.

The only thing you need to know is that you want to split this up into two parts and use the equations below.

v = v0 + a*t

y = v0*t + .5*a*t^2

where v = velocity, v0 = initial velocity, a = acceleration (only acceleration here is gravity), and t = time.

I won't do your HW for you. However, for the first part consider only the path of the rock from when it was initially thrown up to when it passes the edge of the cliff. Then, do it for falling from the cliff edge to the ground.

Answer 3

v = gt + vo (Eq 1)
s = 1/2 gt^2 + vo t + so (Eq 2)

where v = velocity; g = acceleration = -9.8 m/s/s; t = time; vo = initial velocity = 12 m/s; s = displacement; so = initial displacement = 70 m.

1. 0 = 1/2 (-9.8) t^2 + 12 t + 70 (Eq 3 from Eq 2)
4.9 t^2 - 12 t - 70 = 0
t = [12 +/- sqrt(12^2 + 4*4.9*70)] / 9.8
t = [12 +/- sqrt(1516)] / 9.8
t = 5.198 sec ==> t = 5.2 sec rounded off (answer a)

2. v = -9.8 * 5.198 + 12 = -38.94 m/s ==> v = -38.9 m/s rounded off (answer b, from Eq 1)

3. To get this, you need to know how high the stone went. Use Eq 1:

0 = -9.8 t + 12
t = 12/9.8 = 120/98 = 60/49 sec. (that's when it starts dropping)
(The velocity is zero at the top.)

Now use Eq 2:

s = 1/2 (-9.8)(12/9.8)^2 + 12(12/9.8) + 70

s = - 72/9.8 + 144/9.8 + 70 = 72/9.8 + 70 = 7.347 + 70

What this means is that the stone went up 7.347 meters, then fell 77.347 meters. The total distance traveled was 84.7 meters (answer c).

Answer 4

without the size of the stone ( and density ) : the terminal velocity (b) time (a) and distance (c) will be unknown ( if this is a school question it shows why students graduate lacking some basic knowledge )

air resistance is the missing factor - it will effect how far up it goes and how fast it comes down
your teacher should have said a vacuum why don't you take this in to show why the other answers you get are wrong !

Answer 5

I took 2 physics classes on the 1st 3 hundred and sixty 5 days college point and accomplished A's in the two. the class standard grew to become into someplace interior the C's. in case you ask distinctive human beings, you will get distinctive solutions. I did nicely because of the fact I savour physics and did college point issues whilst i grew to become into in highschool.

Answer 6

It will never hit the bottom. It will come back to the cliff