A projectile is launched off a 71m high cliff with and lands 71m away. What was the initial velocity?

Initial velocity = 18.68  m/s

Explanation:

If the projectile is launched horizontally, then its vertical motion is that of free fall (no initial velocity in the vertical direction) and we can estimate the time of its flight via the free fall equation:

y = - 1/2  g  t^2

- 71 = - 4.9  t^2

t^2 = 71/4.9

t = 3.8 s

Now, knowing the time of flight, we can estimate the horizontal launching velocity:

x = v  t

71 = v (3.8)

v = 71/ 3.8

v = 18.68  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

i think the answer would be reserve.