Avery strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 12.0 m/s. how long does he have to get out of the way if the shot was released at a height of 2.10 m, and he is 1.90 m tall?

The shot has a constant downward acceleration of . Since acceleration is constant, the velocity of the ball is given by

where is the shot's vertical velocity at time and is its initial vertical velocity. We're given that . So

gives the shot's velocity at time .

Because acceleration is constant, we also know that

where is the average vertical velocity, is the vertical displacement at time , and is the initial displacement at . We're given an initial displacement of , so the displacement of the shot is

The shotputter is 1.90 m tall, a difference of 0.20 m from the initial height of the throw. So we want to find the time for the ball to reach a displacment of -0.20 m:

  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

an energy transformation flow diagram is shown

where is it ?