Amountain climber stands at the top of a 50 meter cliff hanging over a calm pool of water. the climber throws two stone vertically 1 second apart and observes that they cause a single splash when they hit the water. the first stone has a initial velocity of +2.0 meters per second. what will the velocity of each stone be at the instant bitg stones hit the water?

a) First stone, Vf₁ = 31.38 m/s

b) Second stone, Vf₂ = 34.81 m/s

Explanation:

1) Data:

First stone:

Vo₁ = + 2 m/s (positive sign is understood as throw in the same direction of the acceleration, i.e. downward).

h = 50 m

t₁ = t

Vf₁ = ?

Second stone:

t ₂ = t - 1

Vf₂ = ?

2) Solution

i) First stone, final velocity:

Vf² = Vo² + 2gh

Vf² = (2 m/s)² + 2(9.81m/s²)(50m) = 984 m²/s²

Vf = 31.38 m/s

ii) First stone time, t

Vf = Vo + gt ⇒ t = [Vf - Vo] / g = [31.38 m/s - 2m/s] / 9.81 m/s² = 3.0 s

iii) Second stone time, t - 1

t₂ = t - 1 = 3.0s - 1.0s = 2.0s

iv) Second stone initial velocity, Vo₂

h = Vo₂ t + gt₂² / 2 ⇒ 50m = Vo₂ (2.0s) + 9.81m/s² (2.0s)² / 2

Vo₂ = [ 50m - 19.62m] / (2.0s) = 15.19 m/s

v) Second stone final velocity, Vf₂

Vf₂ = Vo₂ + gt₂ = 15.19 m/s + 9.81 m/s² (2.0s) = 34.81 m/s

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

there are no contact forces mentioned on the list of choices that you've provided.