A stone is thrown verticall upward

[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 + volt + 1/2 at2
Δy = volt + 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= volt + 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!

 

[Gnosis]

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.

 

[ironhawk316]

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

 

[tiny-tim]

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!)

 

[Gnosis]

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.

 

[mplayer]

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.

 

[kalilo]

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~

 

[nikhil.philip]

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

 

[tiny-tim]

welcome to pf!

hi nikhil.philip! welcome to pf!

… wat is its acceleration while it is going up!

you mean after the person let's go of it?

ok, as usual we can use good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" …

first, what are the forces acting on it? ​