Apair of glasses is dropped from the top of a 42.0 meter high stadium. a pen is dropped from the same position 2.00 seconds later. how high above the ground is the pen when the glasses hit the ground? (disregard air resistance. a = - g = - 9.81 m/s ^2)

Given that,

For glasses:

distance, s = 42 m

a= -g = -9.81 m/s²

we can calculate time taken by glasses to hit the ground as follow:

s= Vi* t + 1/2 at²

Since at top initial velocity, Vi= 0

42 = 0 - 0.5* 9.81 t²

t = 2.926 s

Pen is drop 2 second after the glasses is dropped. So time difference will be:

Δt = 2.926 - 2 = 0.926 s

Distance covered by Pen in 0.926 s can be calculated as:

s = 0 - 0.5* 9.81 * 0.926²

s= 4.2 m

Distance of pen from ground will be, 42-4.2 = 37.8 m

When the glasses hit the ground, the pen will be 37.8 m above the ground.

  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