019 10.0 points
a ball is thrown horizontally from the top of
a building 120 m high. the ball strikes the
ground 64 m horizontally from the point of
release.
what is the speed of the ball just before it
strikes the ground?
answer in units of m/s.
020 (ment 1
210,

50.2 m/s

Explanation:

First of all, we need to find the time it takes for the ball to reach the ground.

The vertical position of the ball at time t is given by

where

h = 120 m is the initial height

u = 0 is the initial vertical velocity

g = -9.8 m/s^2 is the acceleration of gravity

The ball reaches the ground when y = 0. Substituting into the equation and solving for t, we find the time of light:

The vertical component of the velocity of the ball changes following the equation

Substituting t = 4.95 s, we find the final vertical velocity of the ball just before reaching the ground:

where the negative sign means the direction is downward.

We also can find the horizontal component of the velocity: since we know the horizontal distance travelled is d = 64 m,

And the final speed is calculated as the magnitude of the resultant of the two components of the velocity:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

b.

hope this : )