A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i. e. its speed).

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is  

v' =  u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

  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