# A stone is thrown verticall upward

• ironhawk316
In summary: Since there is no stated angle of trajectory, there are no known forces acting on the stone. Without knowing the forces acting on the stone, it is impossible to calculate its acceleration.
ironhawk316

## Homework Statement

Question Details:
9. Problem: A stone is thrown verticall upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff? What is its speed just before hitting the ground? What total distance did it travel?

time=
speed=
distance=

## Homework Equations

y = yo + vot + 1/2 at2
Δy = Vot + 1/2 gt2
v2 = vo2 + 2g Δy
v = vo + gt

## The Attempt at a Solution

v = vo + gt
0=12 + (-9.8)t
-12= -9.8t
1.22=t

V2 = Vo2 +2g Δy
02 = (12)2 +2 (-9.8) Δy
02 = 144 + (-19.6) Δy
-144 = -19.6 Δy
7.35=Δy

Δy= Vot + 1/2 gt2
77.35 = o +1/2 (9.8)t2
77.35 = o +4.9 t2
15.79 = t2
3.97 = t

1.22 t + 3.97 t = 5.19 t

distance is 7.35m x 2 + 70m = 84.7 m

V= Vo = at
= -12 + 9.8 (5.2)
= -12 + 50.96
= 38.9 m/s

am I close or totally lost? Help please!

ironhawk316 said:

## Homework Statement

Question Details:
9. Problem: A stone is thrown verticall upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff? What is its speed just before hitting the ground? What total distance did it travel?

"Vertically upward", as stated in your scenario, fails to define any deviation from vertical therefore, there would be no reason for the stone to move in the horizontal direction. This would prevent the stone from ever falling over the 70 meter cliff. Your scenario is missing an angle of trajectory.

That is the whole problem. Where do I go next then or what is the right equation?

Hi ironhawk316!

Why are you doing this in stages?

Use s = v0t + at2/2 starting at the start and finishing at the finish.

(oh … and please don't use yellow!)

ironhawk316 said:
That is the whole problem. Where do I go next then or what is the right equation?

Without an angle of trajectory provided, there won't be any free-falling of the stone over the 70 meter cliff. Since they want the values as per the 70 meter free-fall over the cliff, they must provide the angle of trajectory for the stone.

"Vertical upward" or "vertically upward" fails to provide any X (horizontal) direction and they fail to provide any deviation from vertical. The stone will simply fall right back to where it was thrown unless they provide you with an angle of trajectory. The scenario stated is lacking crucial info.

Maybe the stone-thrower's arm is extended over the edge of the cliff. I think that might be what the problem wants you to assume.

why don't you use v^2 - u^2 = 2as directly after getting displacement?
s = displacement
u = initial velocity
v = final velocity
a = gravity = -9.8

getting displacement = 12 x 1.2
= 14.4m
single vertical distance of rock = 7.2m

s = 77.2
u = 0
v = ?
a = -9.8
t = ?

then directly use v^2 - u^2 = 2as
v^2 - 0 = 2(-9.8)(77.2)
v^2 - 0 = 2(-9.8)(77.2)
v^2 = 1513.12
v = 38.9m/s

this should be easier than what you did above~

i need help!
a stone is thrown upward in a vertical manner..
wat is its acceleration while it is going up!

Last edited by a moderator:

## What is the definition of a stone being thrown vertically upward?

A stone being thrown vertically upward refers to an object being propelled into the air with an initial upward velocity and then falling back down to the ground due to the force of gravity.

## What factors affect the stone's upward motion?

The stone's initial velocity, mass, and the force of gravity are the main factors that affect its upward motion.

## How does the stone's upward motion change over time?

The stone's upward motion initially slows down due to the force of gravity, reaches a maximum height, and then falls back down to the ground at an increasing speed due to the acceleration of gravity.

## What is the equation for calculating the stone's maximum height?

The equation for calculating the stone's maximum height is h = (v^2)/(2g), where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity (9.8 m/s^2).

## How does the angle at which the stone is thrown affect its upward motion?

The angle at which the stone is thrown affects its upward motion by determining the initial velocity and the direction of the motion. A steeper angle will result in a higher initial velocity in the vertical direction and a longer flight time, while a shallower angle will result in a lower initial velocity and a shorter flight time.

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